Probability Rules and Random Variables

Sample Spaces, Outcomes, and Events are the foundation of probability. Before we can calculate how likely something is, we must first describe all possible results of a random experiment. Once we know the full set of possible outcomes, we can identify the outcomes we care about and calculate probabilities correctly.

In this lesson, you will learn how to:

• identify random experiments,
• list the sample space,
• describe individual outcomes,
• define events as sets of outcomes,
• use tree diagrams, tables, and set notation to organize outcomes.

This idea is important because many probability mistakes happen when students leave out outcomes, count outcomes twice, or describe events too vaguely.

1. Random experiments

A random experiment is a process with a result that cannot be predicted with certainty ahead of time, even if all possible results are known.

Examples of random experiments include:

• flipping a coin,
• rolling a die,
• drawing a card from a deck,
• choosing a student at random from a class.

Even though the exact result is unknown before the experiment happens, the set of possible results can still be described.

2. Outcomes

An outcome is a single possible result of a random experiment.

For example:

• When a coin is flipped once, possible outcomes are H or T.
• When a die is rolled once, possible outcomes are 1, 2, 3, 4, 5, 6.
• When two coins are flipped, outcomes are HH, HT, TH, TT.

Each of these individual results is one outcome.

3. Sample space

The sample space is the set of all possible outcomes of a random experiment. It is usually written using the symbol \(S\) or sometimes \(\Omega\).

Examples:

• One coin flip: \(S = \{H, T\}\)
• One die roll: \(S = \{1,2,3,4,5,6\}\)
• Two coin flips: \(S = \{HH, HT, TH, TT\}\)

A correct sample space must include every possible outcome exactly once.

4. Events

An event is a set of one or more outcomes from the sample space. In other words, an event is a collection of outcomes that match a condition.

For example, if a die is rolled and

\(S = \{1,2,3,4,5,6\}\),

then:

• the event “roll an even number” is \(E = \{2,4,6\}\),
• the event “roll a number greater than 4” is \(F = \{5,6\}\),
• the event “roll a 3” is \(G = \{3\}\).

An event can contain:

• one outcome — called a simple event,
• several outcomes — called a compound event.

5. Set notation for sample spaces and events

Because sample spaces and events are sets, set notation is very useful.

• Curly brackets show a set: \(\{\}\)
• If \(A\) is an event, then \(A \subseteq S\), meaning every outcome in \(A\) is in the sample space.
• The empty set \(\varnothing\) represents an impossible event.
• The whole sample space \(S\) represents a certain event.

Example with one die roll:

\(S = \{1,2,3,4,5,6\}\)

Let:

• \(A = \{2,4,6\}\) be the event “even”
• \(B = \{4,5,6\}\) be the event “greater than 3”

Then:

• \(A \cap B = \{4,6\}\) means outcomes in both events
• \(A \cup B = \{2,4,5,6\}\) means outcomes in at least one of the events

These ideas are important later when working with probability rules.

6. Ways to represent a sample space

There are several common ways to organize outcomes clearly.

a) Listing outcomes

This works well when the number of outcomes is small.

Example: rolling one die

\(S = \{1,2,3,4,5,6\}\)

b) Tree diagrams

A tree diagram is helpful when an experiment happens in stages, such as flipping two coins or choosing items one after another.

Each branch represents a possible result at each stage. By following the branches from start to finish, we can list all final outcomes.

c) Tables or charts

Tables are useful when two experiments happen together, such as rolling two dice. One die can label the rows and the other die can label the columns.

This helps make sure no outcomes are missed.

7. Worked Example 1: One die roll

Experiment: Roll a fair six-sided die once.

Step 1: Write the sample space.

$$S = \{1,2,3,4,5,6\}$$

Step 2: Identify outcomes.

Each number is one outcome.

Step 3: Define events.

• Event \(A\): roll an odd number
• Event \(B\): roll a number less than 4

So:

$$A = \{1,3,5\}$$

$$B = \{1,2,3\}$$

Step 4: Find outcomes in both events.

$$A \cap B = \{1,3\}$$

This means the outcomes that are both odd and less than 4 are 1 and 3.

8. Worked Example 2: Two coin flips using a tree idea

Experiment: Flip a coin twice.

For the first flip, the possibilities are \(H\) or \(T\). For each of those, the second flip can also be \(H\) or \(T\).

The sample space is:

$$S = \{HH, HT, TH, TT\}$$

Now define some events:

• Event \(A\): exactly one head
• Event \(B\): at least one tail

Then:

$$A = \{HT, TH\}$$

$$B = \{HT, TH, TT\}$$

Notice that \(HH\) is not in event \(B\) because it has no tails.

If we want the outcomes that are in both events, we find:

$$A \cap B = \{HT, TH\}$$

That makes sense because outcomes with exactly one head automatically have at least one tail.

9. Worked Example 3: Rolling two dice with an ordered sample space

Experiment: Roll two six-sided dice.

Now each outcome must show the result of the first die and the second die. So outcomes are ordered pairs.

Examples of outcomes include:

• \((1,1)\)
• \((2,5)\)
• \((6,3)\)

The full sample space is:

$$S = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\ldots,(6,6)\}$$

There are:

$$6 \times 6 = 36$$

possible outcomes.

Let event \(A\) be “the sum is 7.” Then:

$$A = \{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$$

Let event \(B\) be “both dice show the same number.” Then:

$$B = \{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$$

Notice that order matters in the sample space. For example, \((2,5)\) and \((5,2)\) are different outcomes.

This is a very common place where students make mistakes.

10. Worked Example 4: Selecting from a set and writing events clearly

Experiment: A card is chosen at random from the set \(\{1,2,3,4,5,6,7,8,9,10\}\).

The sample space is:

$$S = \{1,2,3,4,5,6,7,8,9,10\}$$

Define these events:

• \(A\): selecting a prime number
• \(B\): selecting an even number

The prime numbers in the sample space are:

$$A = \{2,3,5,7\}$$

The even numbers are:

$$B = \{2,4,6,8,10\}$$

The event “prime and even” is:

$$A \cap B = \{2\}$$

The event “prime or even” is:

$$A \cup B = \{2,3,4,5,6,7,8,10\}$$

This example shows that events can overlap. An outcome can belong to more than one event.

11. Important ideas to remember

• A sample space contains every possible outcome.
• An outcome is one single result.
• An event is a set of outcomes.
• Events must come from the sample space, so every event is a subset of \(S\).
• For multi-step experiments, use a tree diagram or table to avoid missing outcomes.
• When order matters, write outcomes carefully, such as \((2,5)\) versus \((5,2)\).

12. Common mistakes

• Leaving out outcomes: For two coin flips, writing only \(\{HH, HT, TT\}\) is incorrect because \(TH\) is missing.
• Combining different outcomes: For two dice, \((2,5)\) and \((5,2)\) are different unless the question specifically says order does not matter.
• Describing events too loosely: Instead of saying “high number,” define the event clearly, such as \(\{4,5,6\}\).
• Confusing an outcome with an event: A single outcome like \(3\) can be an event, but an event can also include several outcomes.

13. Quick check questions

1. If a coin is flipped once, what is the sample space?
2. If a die is rolled once, what event represents getting a number greater than 2?
3. When two coins are flipped, why are \(HT\) and \(TH\) different outcomes?
4. When rolling two dice, how many outcomes are in the sample space?

Answers:

1. \(\{H,T\}\)
2. \(\{3,4,5,6\}\)
3. Because the order of the flips matters: first flip then second flip.
4. \(36\)

Summary

To solve probability problems well, start by identifying the sample space. The sample space is the full set of possible outcomes. An outcome is one result, and an event is a group of outcomes that meet a condition.

Tree diagrams, tables, and set notation help organize outcomes clearly. If your sample space is complete and accurate, then finding probabilities becomes much easier and more reliable.

Set Theory in Probability helps us describe events clearly and combine probabilities correctly. In probability, an event is a set of outcomes from an experiment. For example, when rolling a die, the event “getting an even number” is the set \(\{2,4,6\}\).

This lesson focuses on how set ideas such as union and intersection are used in probability, especially when finding the probability that at least one of two events happens. A common mistake is to add probabilities without checking whether the events overlap. Set theory helps us avoid that mistake.

By the end of this lesson, you should be able to:

• describe events using set notation,
• understand unions and intersections of events,
• use the addition rule correctly,
• handle both mutually exclusive and non-mutually exclusive events,
• solve probability problems using Venn diagram thinking.

1. Events as sets

In set theory, a set is a collection of objects. In probability, those objects are outcomes. The full set of all possible outcomes is called the sample space, usually written as \(S\).

For example, if a standard die is rolled, then the sample space is

$$S=\{1,2,3,4,5,6\}.$$

If we define:

• \(A\): getting an even number \(=\{2,4,6\}\)
• \(B\): getting a number greater than 3 \(=\{4,5,6\}\)

then \(A\) and \(B\) are events, and we can compare them using set operations.

2. Important set operations in probability

There are two especially important set ideas in this topic.

• Union: \(A \cup B\) means outcomes in \(A\) or \(B\) or both.
• Intersection: \(A \cap B\) means outcomes that are in both \(A\) and \(B\).

Using the die example:

• \(A=\{2,4,6\}\)
• \(B=\{4,5,6\}\)
• \(A \cup B=\{2,4,5,6\}\)
• \(A \cap B=\{4,6\}\)

The union includes everything in either event. The intersection includes only the outcomes shared by both events.

3. Mutually exclusive and overlapping events

Two events are mutually exclusive if they cannot happen at the same time. In set language, that means their intersection is empty:

$$A \cap B = \varnothing.$$

For example, on one roll of a die:

• \(A\): get a 2
• \(B\): get a 5

These cannot happen together, so they are mutually exclusive.

But many events are not mutually exclusive. They overlap. For example:

• \(A\): get an even number
• \(B\): get a number greater than 3

These overlap at \(\{4,6\}\), so they are not mutually exclusive.

This distinction matters because it changes how we add probabilities.

4. The addition rule for probabilities

If you want the probability that event \(A\) or event \(B\) happens, you are finding \(P(A \cup B)\).

The general addition rule is:

$$P(A \cup B)=P(A)+P(B)-P(A \cap B).$$

This formula is very important. We subtract \(P(A \cap B)\) because outcomes in both events get counted twice when we add \(P(A)\) and \(P(B)\).

Here is the idea:

• \(P(A)\) counts everything in \(A\)
• \(P(B)\) counts everything in \(B\)
• the overlap \(P(A \cap B)\) is counted in both, so subtract it once

If \(A\) and \(B\) are mutually exclusive, then \(P(A \cap B)=0\). The formula becomes:

$$P(A \cup B)=P(A)+P(B).$$

5. Why overlap must be subtracted

Suppose a class survey shows:

• 60% study Mathematics,
• 45% study Physics,
• 25% study both Mathematics and Physics.

If we simply add \(0.60+0.45=1.05\), we get 105%, which is impossible. The problem is that students who study both subjects were counted twice.

So we use the addition rule:

$$P(M \cup P)=0.60+0.45-0.25=0.80.$$

So 80% of students study Mathematics or Physics or both.

6. Visualising with Venn diagrams

A Venn diagram is a helpful way to think about probability sets. Each circle represents an event. The overlapping part represents the intersection.

When solving questions:

1. write down what each event means,
2. identify whether the events overlap,
3. if they overlap, find the intersection,
4. use the addition rule carefully.

A useful way to organize values in a Venn diagram is:

• put the intersection in first,
• then fill in the “A only” and “B only” regions,
• then check totals.

7. Worked Example 1: Simple union with overlap

A fair die is rolled once.

• Let \(A\) be the event “the number is even”.
• Let \(B\) be the event “the number is greater than 3”.

Find \(P(A \cup B)\).

Step 1: List the events.

• \(A=\{2,4,6\}\)
• \(B=\{4,5,6\}\)

Step 2: Find the intersection.

$$A \cap B=\{4,6\}.$$

So:

• \(P(A)=\frac{3}{6}=\frac{1}{2}\)
• \(P(B)=\frac{3}{6}=\frac{1}{2}\)
• \(P(A \cap B)=\frac{2}{6}=\frac{1}{3}\)

Step 3: Use the addition rule.

$$P(A \cup B)=P(A)+P(B)-P(A \cap B).$$

$$P(A \cup B)=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3}.$$

Answer: \(P(A \cup B)=\frac{2}{3}\).

You can also check by listing the union:

$$A \cup B=\{2,4,5,6\},$$

which has 4 outcomes, so \(\frac{4}{6}=\frac{2}{3}\).

8. Worked Example 2: Mutually exclusive events

A card is drawn from a standard deck of 52 cards.

• Let \(A\) be the event “the card is a king”.
• Let \(B\) be the event “the card is a queen”.

Find \(P(A \cup B)\).

Step 1: Find each probability.

There are 4 kings and 4 queens, so

$$P(A)=\frac{4}{52}, \qquad P(B)=\frac{4}{52}.$$

Step 2: Check whether the events overlap.

A single card cannot be both a king and a queen, so the events are mutually exclusive.

$$P(A \cap B)=0.$$

Step 3: Apply the addition rule.

$$P(A \cup B)=P(A)+P(B)-P(A \cap B).$$

$$P(A \cup B)=\frac{4}{52}+\frac{4}{52}-0=\frac{8}{52}=\frac{2}{13}.$$

Answer: \(P(A \cup B)=\frac{2}{13}\).

This is the simpler case because there is no overlap to subtract.

9. Worked Example 3: Using given probabilities

In a group of students:

• \(P(A)=0.55\) where \(A\) is the event “student plays a sport”
• \(P(B)=0.40\) where \(B\) is the event “student plays a musical instrument”
• \(P(A \cap B)=0.18\)

Find the probability that a student plays a sport or a musical instrument.

Step 1: Write the formula.

$$P(A \cup B)=P(A)+P(B)-P(A \cap B).$$

Step 2: Substitute values.

$$P(A \cup B)=0.55+0.40-0.18=0.77.$$

Answer: The probability is \(0.77\).

This means there is a 77% chance that a randomly chosen student does at least one of the two activities.

10. Worked Example 4: Finding the intersection from the union

Suppose:

• \(P(A)=0.62\)
• \(P(B)=0.35\)
• \(P(A \cup B)=0.80\)

Find \(P(A \cap B)\).

Step 1: Start with the addition rule.

$$P(A \cup B)=P(A)+P(B)-P(A \cap B).$$

Step 2: Substitute known values.

$$0.80=0.62+0.35-P(A \cap B).$$

$$0.80=0.97-P(A \cap B).$$

Step 3: Solve.

$$P(A \cap B)=0.97-0.80=0.17.$$

Answer: \(P(A \cap B)=0.17\).

This tells us 17% of outcomes are in both events.

11. Common mistakes to avoid

• Adding probabilities without checking overlap. If events are not mutually exclusive, you must subtract the intersection.
• Confusing union and intersection. Union means “or”; intersection means “and”.
• Forgetting that “or” in probability usually means inclusive or. That means one event, the other event, or both.
• Using the mutually exclusive rule in the wrong situation. Only use \(P(A \cup B)=P(A)+P(B)\) when \(A \cap B=\varnothing\).

12. Key ideas to remember

• An event is a set of outcomes.
• \(A \cup B\) means outcomes in \(A\) or \(B\) or both.
• \(A \cap B\) means outcomes in both \(A\) and \(B\).
• If events overlap, subtract the overlap once.
• The general addition rule is:

$$P(A \cup B)=P(A)+P(B)-P(A \cap B).$$

Brief Summary

Set theory gives a clear way to describe events in probability. The most important rule in this topic is the addition rule for unions: \(P(A \cup B)=P(A)+P(B)-P(A \cap B)\). If events are mutually exclusive, the intersection is zero, so you simply add. If they overlap, you must subtract the overlap so outcomes are not counted twice.

Conditional Probability is used when we want to find the chance of an event happening given that another event has already happened.

This is important because once we know one event has occurred, the sample space changes. We no longer look at all possible outcomes. Instead, we only look at the outcomes that fit the given information.

For example, if you pick a card from a deck and learn that the card is a heart, the probability that it is a queen is no longer based on all 52 cards. It is now based only on the 13 hearts. That is the main idea behind conditional probability.

Notation: The probability of event \(A\) happening given that event \(B\) has happened is written as \(P(A\mid B)\). This is read as “the probability of \(A\) given \(B\).”

The formula for conditional probability is:

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

This formula only works when \(P(B) > 0\).

Here:

• \(P(A\mid B)\) means the probability of \(A\) given \(B\)
• \(P(A\cap B)\) means the probability that both \(A\) and \(B\) happen
• \(P(B)\) means the probability that \(B\) happens

Why does this formula make sense?

If we already know that \(B\) happened, then event \(B\) becomes our new “whole.” Out of all outcomes in \(B\), we count how many also belong to \(A\). That is why we divide by \(P(B)\).

You can think of it as:

$$\text{conditional probability} = \frac{\text{part of }B\text{ that is also }A}{\text{all of }B}$$

Visual idea using sets:

If events \(A\) and \(B\) overlap, then \(A\cap B\) is the overlap. Conditional probability asks: if we are restricted to event \(B\), what fraction of \(B\) lies in that overlap?

Important connection to probability rules

From the conditional probability formula, we can rearrange to get:

$$P(A\cap B)=P(A\mid B)P(B)$$

We can also write:

$$P(A\cap B)=P(B\mid A)P(A)$$

These forms are very useful when solving problems involving “and” probabilities.

When events are independent

Two events \(A\) and \(B\) are independent if knowing that one happened does not change the probability of the other.

In that case:

$$P(A\mid B)=P(A)$$

and

$$P(B\mid A)=P(B)$$

Also, for independent events:

$$P(A\cap B)=P(A)P(B)$$

Be careful: independent does not mean mutually exclusive.

• Independent: one event does not affect the other
• Mutually exclusive: the two events cannot happen together, so \(P(A\cap B)=0\)

If two events are mutually exclusive and \(P(B)>0\), then:

$$P(A\mid B)=0$$

because once \(B\) has happened, \(A\) cannot happen.

Worked Example 1: Rolling a die

A fair six-sided die is rolled once. Let:

• \(A\): the result is even
• \(B\): the result is greater than 3

Find \(P(A\mid B)\).

Step 1: List event \(B\)

If the result is greater than 3, the possible outcomes are \(\{4,5,6\}\).

Step 2: From those outcomes, find which are also in \(A\)

The even outcomes are \(\{2,4,6\}\). In \(B\), the outcomes that are also even are \(\{4,6\}\).

Step 3: Compute the conditional probability

Out of the 3 outcomes in \(B\), 2 satisfy \(A\).

$$P(A\mid B)=\frac{2}{3}$$

Check using the formula:

$$P(A\cap B)=P(\{4,6\})=\frac{2}{6}=\frac{1}{3}$$

$$P(B)=\frac{3}{6}=\frac{1}{2}$$

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1/3}{1/2}=\frac{2}{3}$$

Worked Example 2: Drawing from a deck

A standard deck has 52 cards. One card is drawn. Let:

• \(A\): the card is a queen
• \(B\): the card is a heart

Find \(P(A\mid B)\).

Step 1: Restrict the sample space to hearts

Since we are given that the card is a heart, we only consider the 13 hearts.

Step 2: Count how many hearts are queens

There is exactly one queen of hearts.

So:

$$P(A\mid B)=\frac{1}{13}$$

Check using the formula:

$$P(A\cap B)=\frac{1}{52}$$

$$P(B)=\frac{13}{52}=\frac{1}{4}$$

$$P(A\mid B)=\frac{1/52}{1/4}=\frac{1}{13}$$

This example shows how the given information changes the denominator. We no longer divide by 52, because once we know the card is a heart, only 13 outcomes matter.

Worked Example 3: Two students and a survey

In a class survey:

• 60% of students study mathematics after school
• 40% play a sport after school
• 25% do both

Find the probability that a student plays a sport given that the student studies mathematics after school.

Let:

• \(M\): studies mathematics after school
• \(S\): plays a sport after school

We want \(P(S\mid M)\).

Use the formula:

$$P(S\mid M)=\frac{P(S\cap M)}{P(M)}$$

Substitute the values:

$$P(S\mid M)=\frac{0.25}{0.60}=\frac{25}{60}=\frac{5}{12}$$

So:

$$P(S\mid M)=\frac{5}{12}\approx 0.417$$

Interpretation: Among students who study mathematics after school, about 41.7% also play a sport.

Worked Example 4: Medical testing and Bayes’ theorem

Conditional probability is also the basis of Bayes’ theorem, which helps us reverse a condition.

Suppose:

• 2% of people in a population have a certain condition
• If a person has the condition, the test is positive 95% of the time
• If a person does not have the condition, the test is positive 10% of the time

Find the probability that a person has the condition given that the test is positive.

Let:

• \(C\): person has the condition
• \(T\): test is positive

We want \(P(C\mid T)\).

Step 1: Identify the given probabilities

$$P(C)=0.02$$

$$P(T\mid C)=0.95$$

$$P(T\mid C^c)=0.10$$

Also,

$$P(C^c)=1-0.02=0.98$$

Step 2: Find \(P(T)\)

A positive test can happen in two ways:

• the person has the condition and tests positive
• the person does not have the condition and still tests positive

So:

$$P(T)=P(T\cap C)+P(T\cap C^c)$$

Using conditional probability:

$$P(T)=P(T\mid C)P(C)+P(T\mid C^c)P(C^c)$$

$$P(T)=(0.95)(0.02)+(0.10)(0.98)$$

$$P(T)=0.019+0.098=0.117$$

Step 3: Apply Bayes’ theorem

$$P(C\mid T)=\frac{P(T\mid C)P(C)}{P(T)}$$

$$P(C\mid T)=\frac{(0.95)(0.02)}{0.117}$$

$$P(C\mid T)=\frac{0.019}{0.117}\approx 0.162$$

So the probability that a person has the condition given a positive test is about:

$$0.162=16.2\%$$

Why is this not 95%?

Because the condition is rare. Even though the test is good at detecting the condition, false positives from the much larger group without the condition can still make up a big part of all positive results.

Bayes’ theorem

The formula used above is called Bayes’ theorem:

$$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$$

This is especially useful when we know \(P(B\mid A)\) but want to find \(P(A\mid B)\).

How to solve conditional probability problems

1. Identify the event you are asked to find first. This is the event before the bar in \(P(A\mid B)\).
2. Identify the given event. This is the event after the bar.
3. Reduce the sample space using the given event.
4. Use either counting, the formula \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\), or Bayes’ theorem if needed.
5. Check that your answer makes sense and is between 0 and 1.

Common mistakes to avoid

• Mixing up \(P(A\mid B)\) and \(P(B\mid A)\). These are usually not the same.
• Forgetting to change the sample space after being told that an event has already happened.
• Using the formula when \(P(B)=0\), which is not allowed.
• Confusing independent events with mutually exclusive events.

Quick comparison example

Suppose in a deck of cards:

• \(A\): card is a king
• \(B\): card is a face card

Then:

$$P(A\mid B)=\frac{4}{12}=\frac{1}{3}$$

But:

$$P(B\mid A)=1$$

because every king is automatically a face card. This shows clearly that reversing the order changes the meaning.

Key ideas to remember

• Conditional probability means probability under a new condition.
• The given event changes the sample space.
• $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$
• $$P(A\cap B)=P(A\mid B)P(B)$$
• For independent events, \(P(A\mid B)=P(A)\).
• Bayes’ theorem helps reverse conditional probabilities.

Brief Summary

Conditional probability tells us how likely an event is when another event is already known to have happened. The main formula is $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$, which works by restricting the sample space to event \(B\). This idea also connects to independence, probability rules for intersections, and Bayes’ theorem.

Independent Events and the Multiplication Rule

Probability helps us measure how likely an event is to happen. In many problems, we want to know the chance that two or more events happen together. To do that, we use the multiplication rule.

A very important idea is whether events are independent or dependent. If events are independent, the result of one event does not affect the probability of the other. If they are dependent, one event changes the chance of the other happening.

This lesson will show you how to recognize independent events, test independence using probabilities, and use multiplication rules to find joint probabilities in sequence trials.

1. Basic Probability Review

The probability of an event is a number between 0 and 1.

• Impossible event: \(P(E)=0\)

• Certain event: \(P(E)=1\)

• Any event: \(0 \le P(E) \le 1\)

When we write \(P(A)\), we mean the probability that event \(A\) happens.

When we write \(P(A \cap B)\), we mean the probability that both event \(A\) and event \(B\) happen.

2. What Does Independent Mean?

Two events are independent if one event happening does not change the probability of the other event.

In symbols, events \(A\) and \(B\) are independent if

This says: the probability of \(B\) given that \(A\) happened is the same as the ordinary probability of \(B\).

Because this works both ways, independent events also satisfy

A very useful equivalent test for independence is

This is the main rule you will use most often.

3. The Multiplication Rule

The general multiplication rule for any two events is

This rule always works. It says:

• First find the probability of \(A\).

• Then multiply by the probability that \(B\) happens after \(A\) has happened.

If the events are independent, then \(P(B\mid A)=P(B)\). So the rule becomes

This is the multiplication rule for independent events.

4. How to Tell Whether Events Are Independent

To determine whether two events are independent, compare

If they are equal, the events are independent. If they are not equal, the events are dependent.

You can also compare \(P(B\mid A)\) and \(P(B)\). If they are equal, the events are independent.

5. Independence in Sequence Trials

Many probability questions involve a sequence of repeated actions, such as:

• flipping a coin several times,

• rolling a die repeatedly,

• selecting with replacement.

These are often independent trials. Each trial has the same probabilities, and one trial does not affect the next.

For example, if a fair coin is flipped twice:

• Probability of head on first flip: \(\frac12\)

• Probability of head on second flip: \(\frac12\)

Since the flips are independent,

For several independent trials, multiply the probabilities of each step.

If events \(A_1, A_2, A_3, \dots, A_n\) are independent, then

6. Independent vs Dependent Situations

It is important to know when multiplication uses ordinary probabilities and when it uses conditional probabilities.

• Independent: use \(P(A \cap B)=P(A)P(B)\)

• Dependent: use \(P(A \cap B)=P(A)P(B\mid A)\)

A common difference appears in drawing objects from a bag or deck.

• With replacement: independent, because the item is put back and probabilities stay the same.

• Without replacement: dependent, because the first draw changes what remains.

7. Worked Example 1: Two Independent Coin Flips

A fair coin is flipped twice. What is the probability of getting a tail on the first flip and a head on the second flip?

Step 1: Define the events.

• \(A\): first flip is tail, so \(P(A)=\frac12\)

• \(B\): second flip is head, so \(P(B)=\frac12\)

Step 2: The flips are independent, so multiply.

Answer: The probability is \(\frac14\).

8. Worked Example 2: Testing Independence Analytically

Suppose

Are events \(A\) and \(B\) independent?

Step 1: Find \(P(A)P(B)\).

Step 2: Compare with \(P(A \cap B)\).

Since

the events are independent.

Answer: Yes, \(A\) and \(B\) are independent.

9. Worked Example 3: Sequence Trials with a Die

A fair six-sided die is rolled three times. What is the probability of rolling a 6, then an even number, then a number less than 3?

Step 1: Find each probability.

• First roll is 6: \(\frac16\)

• Second roll is even: \(\frac36=\frac12\)

• Third roll is less than 3: outcomes 1 or 2, so \(\frac26=\frac13\)

Step 2: The rolls are independent, so multiply.

Answer: The probability is \(\frac1{36}\).

10. Worked Example 4: Dependent vs Independent Drawing

A bag contains 3 red balls and 2 blue balls.

Part A: With replacement

A ball is drawn, its color is recorded, and then it is put back. Then a second ball is drawn. What is the probability of drawing red and then red?

The probability of red on each draw is \(\frac35\). Because the ball is replaced, the draws are independent.

Part B: Without replacement

Now suppose the first red ball is not put back. What is the probability of drawing red and then red?

Step 1: Probability first ball is red:

Step 2: After drawing one red, 2 red balls remain out of 4 total balls. So

Step 3: Use the general multiplication rule.

Notice that

So these events are not independent when there is no replacement.

11. Common Mistakes to Avoid

• Using multiplication automatically. First ask whether the events are independent or whether you need a conditional probability.

• Confusing “and” with “or”. The multiplication rule is for “and” events, meaning both happen.

• Assuming repeated events are always dependent. Coin flips and die rolls are usually independent because each trial starts fresh.

• Ignoring replacement. In drawing problems, replacement often decides whether events are independent.

• Forgetting to test independence correctly. You must check whether \(P(A \cap B)=P(A)P(B)\).

12. Quick Problem-Solving Guide

1. Identify the events.

2. Decide whether they are independent.

3. If independent, use

4. If dependent, use

5. For several independent trials, multiply all the individual probabilities.

6. Check that your final answer is between 0 and 1.

13. Brief Summary

Independent events are events where one event does not change the probability of the other. For independent events, the probability that both happen is the product of their individual probabilities:

For any two events, the general multiplication rule is

In sequence trials like coin tosses, die rolls, and sampling with replacement, multiplication is often used because the trials are independent. In sampling without replacement, the events are usually dependent, so conditional probability is needed.

Bayes' Theorem helps us find a probability in reverse.

Sometimes we know the chance of an effect happening if a cause is true, and we want to work backwards to find the chance that the cause is true after seeing the effect. Bayes' Theorem gives us a way to do that.

For example, we may know the probability that a medical test is positive when a person has a disease. But if a person gets a positive result, that does not automatically mean they probably have the disease. Bayes' Theorem helps us calculate the actual probability.

This idea is important because in real life, we often observe results first and then try to decide what likely caused them.

1. Review: Conditional Probability

Before learning Bayes' Theorem, recall conditional probability.

The probability of event A happening given that event B has happened is written as \(P(A\mid B)\).

It is calculated by

as long as \(P(B)\neq 0\).

Here:

• \(P(A\mid B)\) means the probability of A given B
• \(P(A\cap B)\) means the probability that A and B both happen
• \(P(B)\) means the probability that B happens

Bayes' Theorem is built from this conditional probability formula.

2. The Bayes' Theorem Formula

Bayes' Theorem connects \(P(A\mid B)\) and \(P(B\mid A)\).

The formula is

This is valid when \(P(B)\neq 0\).

Each part has a meaning:

• \(P(A)\): the prior probability of A, before seeing B
• \(P(B\mid A)\): the probability of observing B if A is true
• \(P(B)\): the total probability of B
• \(P(A\mid B)\): the updated probability of A after observing B

In simple words:

3. Why Bayes' Theorem Matters

A common mistake is to confuse \(P(A\mid B)\) with \(P(B\mid A)\). These are usually not equal.

For example:

• \(P(\text{positive test}\mid \text{disease})\): probability the test is positive if a person has the disease
• \(P(\text{disease}\mid \text{positive test})\): probability a person has the disease if the test is positive

These look similar, but they answer different questions.

Bayes' Theorem helps us move from one to the other.

4. Finding \(P(B)\): Total Probability

Often, the hardest part of Bayes' Theorem is finding \(P(B)\), the total probability of the observed event.

If event A can happen or not happen, then B can occur in two ways:

• B happens when A happens
• B happens when A does not happen

So:

Here \(A^c\) means “not A”.

This is very useful in many Bayes' Theorem problems.

Substituting this into the formula gives:

5. How to Solve Bayes' Theorem Problems

A good step-by-step method is:

1. Decide what event you want to find in reverse, such as \(P(A\mid B)\).
2. Identify the prior probability \(P(A)\).
3. Identify the likelihood \(P(B\mid A)\).
4. Find the total probability \(P(B)\).
5. Substitute into Bayes' formula.
6. Check whether the answer makes sense.

Tree diagrams can also help organize the information.

6. Worked Example 1: A Simple School Club Example

At a school, 40% of students are in the music club. Of the students in the music club, 70% also play a sport. Of the students not in the music club, 50% play a sport.

If a randomly chosen student plays a sport, what is the probability that the student is in the music club?

Step 1: Define events

• Let \(M\) = student is in the music club
• Let \(S\) = student plays a sport

We want \(P(M\mid S)\).

Step 2: List the known probabilities

\(P(M)=0.40\)

\(P(M^c)=0.60\)

\(P(S\mid M)=0.70\)

\(P(S\mid M^c)=0.50\)

Step 3: Find \(P(S)\)

Step 4: Apply Bayes' Theorem

Answer: The probability is about \(0.483\), or 48.3%.

Notice that even though 70% of music club students play a sport, the reverse probability is not 70%.

7. Worked Example 2: Medical Testing

A disease affects 2% of a population. A test is 95% accurate for people who have the disease, meaning

The test also gives a false positive 8% of the time for people who do not have the disease, so

If a person tests positive, what is the probability that they actually have the disease?

Step 1: Define events

• Let \(D\) = person has the disease
• Let \(P\) = test is positive

We want \(P(D\mid P)\).

Step 2: List known values

\(P(D)=0.02\)

\(P(D^c)=0.98\)

\(P(P\mid D)=0.95\)

\(P(P\mid D^c)=0.08\)

Step 3: Find the total probability of a positive test

Step 4: Apply Bayes' Theorem

Answer: The probability is about 19.5%.

This may seem surprising at first. The test is quite accurate, but the disease is rare. Because most people do not have the disease, many positive results can still come from false positives.

8. Worked Example 3: Using Counts Instead of Decimals

A factory has two machines.

• Machine A makes 60% of the items and 3% of its items are defective.
• Machine B makes 40% of the items and 7% of its items are defective.

If an item chosen at random is defective, what is the probability that it came from Machine B?

Step 1: Define events

• Let \(B\) = item came from Machine B
• Let \(D\) = item is defective

We want \(P(B\mid D)\).

Step 2: Write the known probabilities

\(P(B)=0.40\)

\(P(A)=0.60\)

\(P(D\mid B)=0.07\)

\(P(D\mid A)=0.03\)

Step 3: Find \(P(D)\)

Step 4: Apply Bayes' Theorem

Answer: The probability is about 60.9%.

Even though Machine B only makes 40% of the items, a defective item is more likely to have come from Machine B because its defect rate is higher.

9. Worked Example 4: Bayes' Theorem with a Tree Diagram Idea

Suppose 30% of students study French and 70% study Spanish. Of the French students, 80% pass an oral exam. Of the Spanish students, 60% pass the oral exam.

If a student passes the oral exam, what is the probability that the student studies French?

Step 1: Define events

• \(F\): studies French
• \(S\): studies Spanish
• \(P\): passes the oral exam

We want \(P(F\mid P)\).

Step 2: Use the given probabilities

\(P(F)=0.30\)

\(P(S)=0.70\)

\(P(P\mid F)=0.80\)

\(P(P\mid S)=0.60\)

Step 3: Find total probability of passing

Step 4: Find the required probability

Answer: The probability is about 36.4%.

10. Common Mistakes to Avoid

• Mixing up \(P(A\mid B)\) and \(P(B\mid A)\)
• Forgetting to calculate the total probability in the denominator
• Using the wrong prior probability
• Ignoring the complement, such as \(A^c\)
• Assuming a high test accuracy means a high final probability

Always ask yourself: “Am I finding the probability of the cause given the result, or the result given the cause?”

11. A Helpful Interpretation

Bayes' Theorem can be thought of as a way to update what we believe when new information appears.

You start with an initial belief, called the prior probability. Then you observe some evidence. Bayes' Theorem tells you how to combine the old information and the new evidence to get an updated probability.

This is why Bayes' Theorem is often used in science, medicine, decision-making, and data analysis.

12. Quick Formula Box

Main formula:

If needed, find the denominator using:

So another useful form is:

13. Final Summary

Bayes' Theorem is used to reverse conditional probabilities. It helps us find the probability of a cause after observing an effect.

To use it, identify the prior probability, the likelihood, and the total probability of the evidence. Then substitute carefully into the formula.

The key idea is that \(P(A\mid B)\) and \(P(B\mid A)\) are different, and Bayes' Theorem tells us exactly how they are related.

Discrete Random Variables and Distributions

In probability, we often want to describe not just whether an event happens, but what numerical outcome results from a random process. For example, when tossing coins, we might count the number of heads. When rolling a die, we might record the number shown. These numerical outcomes can be organized using a discrete random variable.

A discrete random variable is a variable that takes specific countable values, each with a probability. A probability distribution shows every possible value of the random variable and the probability of each value.

This lesson will show you how to:

• identify a discrete random variable,
• list all possible numerical outcomes,
• calculate the probability of each outcome,
• construct a probability distribution table, and
• check whether a distribution is valid.

1. What is a discrete random variable?

A random variable assigns a number to the outcome of a random experiment. It is usually written with a capital letter such as \(X\).

For example:

• If you roll one die, let \(X\) be the number rolled. Then \(X\) can be \(1,2,3,4,5,6\).
• If you toss three coins, let \(X\) be the number of heads. Then \(X\) can be \(0,1,2,3\).

These are discrete random variables because the possible values are separate, countable numbers.

2. What is a probability distribution?

A probability distribution is a table, list, or rule that gives the probability for each possible value of a random variable.

If \(X\) is a discrete random variable, then its distribution may be written as:

This means “the probability that \(X\) takes the value \(x\).”

A valid probability distribution must satisfy two rules:

• Each probability must be between 0 and 1:

• The probabilities of all possible values must add to 1:

This second rule makes sense because one of the possible outcomes must happen.

3. Steps for constructing a distribution table

When building a distribution table, follow these steps:

1. Describe the random process clearly.
2. Define the random variable \(X\).
3. List all possible values of \(X\).
4. Find the probability of each value.
5. Check that all probabilities add to 1.

It is important to distinguish between outcomes of the experiment and values of the random variable. Several different outcomes may lead to the same value of \(X\).

For instance, in three coin tosses, the outcomes \(HHT\), \(HTH\), and \(THH\) are different experiment outcomes, but they all give \(X=2\) if \(X\) counts the number of heads.

4. Example 1: Rolling one fair die

Question: A fair six-sided die is rolled once. Let \(X\) be the number shown. Construct the probability distribution of \(X\).

Step 1: List possible values.

Because the die has six faces, \(X\) can be:

Step 2: Find probabilities.

Since the die is fair, each value is equally likely:

Distribution table:

\[ \begin{array}{c|cccccc} X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X) & \frac16 & \frac16 & \frac16 & \frac16 & \frac16 & \frac16 \end{array} \]

Check:

So this is a valid probability distribution.

5. Example 2: Tossing two fair coins

Question: Two fair coins are tossed. Let \(X\) be the number of heads. Construct the probability distribution of \(X\).

Step 1: List all outcomes of the experiment.

The sample space is:

Each outcome is equally likely, so each has probability \(\frac14\).

Step 2: Find the possible values of \(X\).

• \(X=0\) if no heads appear: \(TT\)
• \(X=1\) if exactly one head appears: \(HT, TH\)
• \(X=2\) if two heads appear: \(HH\)

So the possible values are:

Step 3: Calculate each probability.

Distribution table:

\[ \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac14 & \frac12 & \frac14 \end{array} \]

Check:

This table shows that getting one head is more likely than getting zero or two heads.

6. Example 3: Sum of two fair dice

Question: Two fair six-sided dice are rolled. Let \(X\) be the sum of the two numbers. Construct the probability distribution of \(X\).

This example is more challenging because different sums can happen in different numbers of ways.

Step 1: Find possible values.

The smallest sum is \(1+1=2\), and the largest sum is \(6+6=12\). So:

Step 2: Count total outcomes.

Each die has 6 outcomes, so the total number of equally likely outcomes is:

Step 3: Count how many ways each sum can occur.

• Sum 2: \((1,1)\) → 1 way
• Sum 3: \((1,2),(2,1)\) → 2 ways
• Sum 4: \((1,3),(2,2),(3,1)\) → 3 ways
• Sum 5: 4 ways
• Sum 6: 5 ways
• Sum 7: 6 ways
• Sum 8: 5 ways
• Sum 9: 4 ways
• Sum 10: 3 ways
• Sum 11: 2 ways
• Sum 12: 1 way

Step 4: Convert counts to probabilities.

Distribution table:

\[ \begin{array}{c|ccccccccccc} X & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(X) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \end{array} \]

Check:

This distribution is not uniform. The sum 7 is the most likely because it can happen in the greatest number of ways.

7. Example 4: A simple non-uniform situation

Question: A bag contains 3 red counters and 2 blue counters. One counter is chosen at random. Let \(X\) be defined by:

• \(X=1\) if a red counter is chosen
• \(X=0\) if a blue counter is chosen

Construct the probability distribution of \(X\).

Step 1: List possible values.

Step 2: Find probabilities.

There are 5 counters in total.

Distribution table:

\[ \begin{array}{c|cc} X & 0 & 1 \\ \hline P(X) & \frac25 & \frac35 \end{array} \]

Check:

This example shows that a probability distribution does not need to have equal probabilities.

8. How to tell if a table is a valid probability distribution

Suppose you are given a table and asked whether it could represent a probability distribution for a discrete random variable. Use these checks:

• Are all probabilities between 0 and 1?
• Do all probabilities add to 1?
• Does each probability match a possible value of the random variable?

For example, this is not valid:

\[ \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & 0.2 & 0.5 & 0.4 \end{array} \]

because:

The probabilities add to more than 1, so the table cannot be a valid probability distribution.

9. Important ideas to remember

• A discrete random variable has countable numerical values.
• A probability distribution lists every possible value and its probability.
• Different experimental outcomes can give the same value of the random variable.
• The probabilities in a valid distribution must add to 1.
• To find probabilities, you often count how many outcomes produce each value.

10. Common mistakes

• Missing possible values: forgetting one of the values the random variable can take.
• Mixing outcomes with variable values: for example, listing \(HT\) as a value of \(X\) when \(X\) is supposed to count heads.
• Incorrect probability totals: not checking that probabilities sum to 1.
• Assuming all values are equally likely: this is not always true, such as with the sum of two dice.

11. Quick method for solving exam-style questions

1. Write what \(X\) represents.
2. List the possible values of \(X\).
3. Use a sample space, counting, or simple fractions to find each probability.
4. Put the results into a clear table.
5. Add the probabilities to confirm the total is 1.

Brief Summary

A discrete random variable gives a numerical value for the result of a random process. Its probability distribution shows each possible value and the probability of getting it. To construct a distribution, list all possible values, calculate each probability carefully, and make sure the probabilities add to 1. This skill is essential for understanding probability models and for working with expected value later on.

Expected Value is the long-run average outcome of a random process. It tells us what we should expect on average if an experiment is repeated many times.

For example, if you play a game many times, the expected value helps predict the average amount you would win or lose per game. It does not mean that one single result must equal the expected value.

Expected value is an important idea in probability because it connects a random variable's possible outcomes with how likely each outcome is.

1. What is a discrete random variable?

A random variable is a variable whose value depends on the outcome of a random event.

• If you roll a fair six-sided die, the random variable could be the number shown: 1, 2, 3, 4, 5, or 6.
• If you count the number of heads in 3 coin tosses, the random variable could be 0, 1, 2, or 3.

A discrete random variable has specific separate values, usually whole numbers.

2. Probability distribution

A probability distribution lists all possible values of a discrete random variable and the probability of each value.

If a random variable is called \(X\), then its expected value is written as \(E(X)\).

For a discrete random variable, the formula is:

This means:

• Multiply each possible outcome \(x\) by its probability.
• Add all those products.

This gives the theoretical mean, or long-run average.

3. Steps for finding expected value

1. List all possible values of the random variable.
2. Write the probability for each value.
3. Multiply each value by its probability.
4. Add the results.

4. Important idea: expected value may not be a possible outcome

The expected value can be a decimal or a number that never actually occurs in one trial.

For example, if you roll a fair die, the expected value is:

You cannot roll a 3.5, but over many rolls, the average result approaches 3.5.

Worked Example 1: Expected value for a fair die

Let \(X\) be the number rolled on a fair six-sided die.

The probability distribution is:

• \(P(X=1)=\frac16\)
• \(P(X=2)=\frac16\)
• \(P(X=3)=\frac16\)
• \(P(X=4)=\frac16\)
• \(P(X=5)=\frac16\)
• \(P(X=6)=\frac16\)

Now calculate:

Answer: The expected value is \(3.5\).

Worked Example 2: Expected value from a probability table

A random variable \(X\) has the following distribution:

• \(P(X=0)=0.2\)
• \(P(X=1)=0.5\)
• \(P(X=2)=0.3\)

Find \(E(X)\).

Step 1: Multiply each value by its probability.

• \(0(0.2)=0\)
• \(1(0.5)=0.5\)
• \(2(0.3)=0.6\)

Step 2: Add the products.

Answer: \(E(X)=1.1\).

This means that over many repetitions, the average value of \(X\) would be 1.1.

5. Expected value in games

Expected value is often used in games involving money. In these questions, the random variable usually represents the amount won, amount lost, or net gain.

Be careful to define the variable clearly. If there is a cost to play a game, you often need the net gain:

Worked Example 3: Expected value of a game

A game costs \(\$2\) to play. A player draws one card from a bag.

• There are 5 cards in total.
• 1 card gives a prize of \(\$10\).
• 4 cards give no prize.

Find the expected net gain for one play.

Step 1: Define the random variable.

Let \(X\) = net gain in dollars.

If the player wins the prize, the net gain is:

If the player does not win, the net gain is:

So the distribution is:

• \(P(X=8)=\frac15\)
• \(P(X=-2)=\frac45\)

Step 2: Calculate the expected value.

Answer: The expected net gain is \(\$0\).

This means that in the long run, the game is fair. On average, a player neither gains nor loses money.

6. Interpreting expected value

The expected value helps us judge whether a situation is favorable.

• If the expected net gain is positive, the game is favorable to the player.
• If the expected net gain is negative, the player is expected to lose money in the long run.
• If the expected net gain is zero, the game is fair.

Worked Example 4: Is the game favorable?

A spinner has three possible prizes:

• Win \(\$12\) with probability \(0.1\)
• Win \(\$4\) with probability \(0.3\)
• Win \(\$0\) with probability \(0.6\)

The game costs \(\$3\) to play. Find the expected net gain and decide whether the game is favorable to the player.

Step 1: Convert prizes to net gains.

• \(12-3=9\)
• \(4-3=1\)
• \(0-3=-3\)

So the random variable \(X\) = net gain has distribution:

• \(P(X=9)=0.1\)
• \(P(X=1)=0.3\)
• \(P(X=-3)=0.6\)

Step 2: Find expected value.

Answer: The expected net gain is \(-\$0.60\).

So in the long run, the player loses an average of 60 cents per game. This game is not favorable to the player.

7. Expected value for counting events

Expected value also appears when a random variable counts how many times something happens.

For example, let \(X\) be the number of heads in 2 coin tosses.

The possible values are 0, 1, and 2.

• \(P(X=0)=\frac14\)
• \(P(X=1)=\frac12\)
• \(P(X=2)=\frac14\)

Then:

So the expected number of heads in 2 tosses is 1.

8. Common mistakes to avoid

• Forgetting to multiply by probability: expected value is not just the average of the outcomes.
• Using prizes instead of net gain: if there is a cost, subtract it first when needed.
• Probabilities not adding to 1: always check that the total probability is 1.
• Thinking expected value must be a possible result: it is often not a value that can happen in a single trial.

9. Quick method checklist

When solving expected value questions, ask yourself:

1. What is the random variable \(X\)?
2. What are all possible values of \(X\)?
3. What is the probability of each value?
4. Am I using outcome, prize, or net gain?
5. Did I calculate \(E(X)=\sum xP(X=x)\) correctly?

Summary

Expected value is the long-run theoretical mean of a discrete random variable. To find it, multiply each possible outcome by its probability and add the results:

This value describes the average result over many repetitions, not necessarily the outcome of one trial. It is especially useful in probability distributions, counting events, and deciding whether games involving money are fair or favorable.

Variance and standard deviation tell us how spread out the values of a random variable are around the mean, or expected value.

In probability, the expected value gives the long-run average outcome, but it does not tell us whether the values usually stay close to that average or vary widely. Two random variables can have the same expected value but very different spreads.

This lesson explains how to find and interpret the variance and standard deviation of a discrete random variable.

1. Review: Discrete random variables and expected value

A random variable assigns a numerical value to each outcome of a chance process.

For a discrete random variable, the possible values are listed, and each value has a probability.

If a random variable \(X\) can take values \(x_1, x_2, x_3, \dots\) with probabilities \(P(X=x_1), P(X=x_2), P(X=x_3), \dots\), then the expected value is

This is the probability-weighted average of all possible values.

2. What variance measures

The variance measures how far the values of a random variable tend to be from the mean.

To do this, we look at each difference from the mean, \(x-\mu\), square it, and then take the probability-weighted average.

The formula for variance is

Important ideas:

• \(x-\mu\) is the deviation from the mean.
• Squaring makes all values positive and gives more weight to values far from the mean.
• Variance is always greater than or equal to 0.

3. What standard deviation measures

The standard deviation is the square root of the variance.

Standard deviation is often easier to interpret because it is measured in the same units as the random variable.

For example, if \(X\) measures number of goals, then variance is in squared goals, but standard deviation is in goals.

4. Why not just average the deviations?

If we simply added the deviations \((x-\mu)\), the positive and negative values would cancel out.

That is why variance uses squared deviations instead.

5. Steps for finding variance and standard deviation

1. Find the mean: \(\mu=E(X)=\sum xP(X=x)\).
2. For each value of \(x\), calculate \((x-\mu)^2\).
3. Multiply each squared deviation by its probability.
4. Add the results to get the variance.
5. Take the square root to get the standard deviation.

6. Shortcut formula for variance

There is a useful shortcut:

where

So you can calculate variance by:

1. Finding \(E(X)\),
2. Finding \(E(X^2)\),
3. Using \(\mathrm{Var}(X)=E(X^2)-[E(X)]^2\).

This method is often faster, especially when there are several values.

7. Worked Example 1: A simple distribution

Suppose a random variable \(X\) has the following distribution:

• \(X=1\) with probability \(0.2\)
• \(X=2\) with probability \(0.5\)
• \(X=4\) with probability \(0.3\)

Step 1: Find the mean

So the mean is \(\mu=2.4\).

Step 2: Find the variance

Use \(\mathrm{Var}(X)=\sum (x-\mu)^2P(X=x)\).

For \(x=1\):

For \(x=2\):

For \(x=4\):

Add them:

Step 3: Find the standard deviation

Answer:

• Mean: \(2.4\)
• Variance: \(1.24\)
• Standard deviation: approximately \(1.11\)

8. Worked Example 2: Using the shortcut formula

A fair die is rolled once. Let \(X\) be the number shown.

The possible values are \(1,2,3,4,5,6\), each with probability \(\frac{1}{6}\).

Step 1: Find \(E(X)\)

Step 2: Find \(E(X^2)\)

Step 3: Use the shortcut formula

Write \(12.25\) as a fraction:

Then

So

Now find the standard deviation:

Answer:

• Mean: \(3.5\)
• Variance: \(\frac{35}{12}\approx 2.92\)
• Standard deviation: approximately \(1.71\)

9. Worked Example 3: Number of heads in two coin tosses

Two fair coins are tossed. Let \(X\) be the number of heads.

The possible values are:

• \(X=0\) with probability \(\frac{1}{4}\)
• \(X=1\) with probability \(\frac{1}{2}\)
• \(X=2\) with probability \(\frac{1}{4}\)

Step 1: Find the mean

So \(\mu=1\).

Step 2: Find the variance

Step 3: Find the standard deviation

Interpretation: The average number of heads is 1, and the results typically differ from 1 by about 0.71 heads.

10. Worked Example 4: Comparing spread

Consider two random variables:

Random variable A

• \(X=4\) with probability \(0.5\)
• \(X=6\) with probability \(0.5\)

Random variable B

• \(X=1\) with probability \(0.5\)
• \(X=9\) with probability \(0.5\)

Step 1: Compare the means

For A:

For B:

Both random variables have the same mean, \(5\).

Step 2: Compare the variances

For A:

So the standard deviation is

For B:

So the standard deviation is

Interpretation: Even though both have mean 5, random variable B is much more spread out. That is why its variance and standard deviation are much larger.

11. Important observations

• If all outcomes are close to the mean, the variance is small.
• If outcomes are far from the mean, the variance is large.
• If a random variable always has the same value, then the variance is 0 and the standard deviation is 0.
• Standard deviation is usually easier to interpret than variance because it uses the original units.

12. Common mistakes to avoid

• Forgetting to find the mean first. You need the mean before using \(\sum (x-\mu)^2P(X=x)\).
• Forgetting to square the deviation. Variance uses \((x-\mu)^2\), not just \(x-\mu\).
• Forgetting to multiply by probability. Every term must be weighted by its probability.
• Mixing up variance and standard deviation. Standard deviation is the square root of variance.
• Using probabilities that do not add to 1. Always check the distribution first.

13. Quick check with a probability table

If a distribution is given in table form, organize your work carefully.

Then:

• Add the \(xP(X=x)\) column to get \(E(X)\).
• Add the \(x^2P(X=x)\) column to get \(E(X^2)\).
• Use \(\mathrm{Var}(X)=E(X^2)-[E(X)]^2\).

14. Final summary

The variance of a discrete random variable measures the average squared distance from the mean:

The standard deviation is the square root of the variance:

You can also use the shortcut formula:

These measures help describe not just the center of a distribution, but also how much the outcomes vary. A small standard deviation means values are close to the mean, while a large standard deviation means the values are more spread out.

The Binomial Distribution

In probability, we often want to find the chance of getting a certain number of successes in a fixed number of repeated trials. For example, we might want the probability of getting exactly 3 correct guesses out of 5 questions, or exactly 7 heads in 10 coin flips. The binomial distribution is the model used for this kind of situation.

This lesson explains when to use the binomial distribution, how its formula works, and how to calculate probabilities step by step.

1. When is a situation binomial?

A random variable follows a binomial distribution if all of these conditions are true:

• There is a fixed number of trials, called \(n\).
• Each trial has only two possible outcomes: success or failure.
• The trials are independent, meaning one trial does not affect another.
• The probability of success is the same each time, called \(p\).

If these four conditions are met, then the number of successes \(X\) is a binomial random variable.

We write this as:

Here:

• \(n\) = number of trials
• \(p\) = probability of success on one trial
• \(X\) = number of successes

2. Examples of binomial settings

• Flipping a coin 8 times and counting the number of heads
• Guessing on 10 true/false questions and counting correct answers
• Testing 12 light bulbs and counting how many are defective, if each bulb has the same defect probability

Not binomial examples:

• Rolling a die and counting the total score, because there are more than two outcomes per roll
• Drawing cards without replacement, because the probabilities change from draw to draw

3. The binomial probability formula

To find the probability of getting exactly \(k\) successes in \(n\) trials, use:

This formula has three important parts:

• \(\binom{n}{k}\): the number of ways to choose which \(k\) trials are successes
• \(p^k\): the probability of getting success \(k\) times
• \((1-p)^{n-k}\): the probability of getting failure in the remaining trials

The combination term is:

The symbol \(!\) means factorial. For example, \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\).

4. Why combinations are needed

Suppose you flip a coin 4 times and want exactly 2 heads. One possible result is HHTT, but that is not the only one. You could also get HTHT, HTTH, THHT, THTH, or TTHH.

All of these outcomes have the same probability if the coin is fair, so we must count how many different arrangements give exactly 2 heads. That is why the combination \(\binom{4}{2}\) appears in the formula.

5. Worked Example 1: A fair coin

A fair coin is flipped 5 times. Find the probability of getting exactly 3 heads.

Step 1: Identify the values

• \(n = 5\)
• \(k = 3\)
• \(p = 0.5\)

Step 2: Use the formula

Because \((0.5)^3(0.5)^2 = (0.5)^5\), we get:

Now calculate the combination:

So:

Answer: The probability of exactly 3 heads is \(0.3125\).

6. Worked Example 2: Multiple-choice guessing

A student guesses on 8 true/false questions. What is the probability the student gets exactly 6 correct?

Each question has:

• success = correct answer
• \(p = 0.5\)
• \(n = 8\)
• \(k = 6\)

Apply the binomial formula:

Now find the combination:

So:

Answer: The probability of exactly 6 correct answers is \(0.109375\).

7. Finding probabilities like “at least” or “at most”

Sometimes a question does not ask for exactly one value. Instead, it may ask for:

• at least \(k\): means \(k\) or more
• at most \(k\): means \(k\) or fewer
• more than \(k\): means greater than \(k\)
• less than \(k\): means smaller than \(k\)

These probabilities are found by adding binomial probabilities.

For example:

Also, the complement rule is often helpful:

8. Worked Example 3: At least one success

A basketball player makes a free throw with probability \(0.7\). She takes 4 shots. Find the probability that she makes at least one shot.

Let \(X\) be the number of successful shots. Then:

• \(n = 4\)
• \(p = 0.7\)

We want:

Instead of adding \(P(X=1)+P(X=2)+P(X=3)+P(X=4)\), it is easier to use the complement:

Now calculate \(P(X=0)\):

So:

Answer: The probability that she makes at least one shot is \(0.9919\).

9. Worked Example 4: At most a certain number

A factory produces batteries. Each battery has a \(0.1\) probability of being defective. A sample of 6 batteries is checked. Find the probability that at most 2 are defective.

Let defective be a success. Then:

• \(n = 6\)
• \(p = 0.1\)

We need:

First, find \(P(X=0)\):

Next, find \(P(X=1)\):

Now, find \(P(X=2)\):

Add them:

Answer: The probability that at most 2 batteries are defective is \(0.98415\).

10. Mean and standard deviation of a binomial distribution

For a binomial random variable \(X \sim \text{Bin}(n,p)\):

These formulas help describe the distribution without finding every probability.

• The mean \(\mu\) is the expected number of successes.
• The standard deviation \(\sigma\) shows how much the number of successes tends to vary.

Example: If \(X \sim \text{Bin}(20, 0.3)\), then:

This means the expected number of successes is 6.

11. Important reminders

• The word success does not mean something good. It just means the outcome you are counting.
• Check carefully that the probability \(p\) stays constant.
• Make sure trials are independent before using a binomial model.
• For “at least” and “at most” questions, be careful about which values to add.
• Use the complement rule when it makes the work shorter.

12. Common mistakes

• Forgetting the combination term \(\binom{n}{k}\)
• Mixing up \(p\) and \(1-p\)
• Using binomial distribution when trials are not independent
• Misreading phrases like “at least” and “at most”

13. How to recognize a binomial question quickly

Ask yourself these four questions:

1. Is there a fixed number of trials?
2. Are there only two outcomes each time?
3. Is the probability of success the same each time?
4. Are the trials independent?

If the answer to all four is yes, then use the binomial distribution.

Brief Summary

The binomial distribution models the number of successes in a fixed number of independent trials when each trial has two outcomes and a constant success probability. If \(X \sim \text{Bin}(n,p)\), then the probability of exactly \(k\) successes is $$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$. For ranges such as “at least” or “at most,” add appropriate probabilities or use the complement rule. The mean is \(np\), and the standard deviation is \(\sqrt{np(1-p)}\).

The Geometric Distribution

When we repeat the same chance process over and over, sometimes we want to know when the first success will happen. For example, how likely is it that the first correct answer happens on the 4th guess? Or that the first made basket happens on the 3rd shot?

The geometric distribution models this situation. It gives the probability that the first success occurs on exactly the \(n\)th trial, when each trial is independent and has the same probability of success.

This distribution is useful in many real-life situations:

• the first time a machine works correctly,
• the first time a student answers a multiple-choice question correctly by guessing,
• the first time a free throw is made,
• the first time a coin lands on heads.

1. Conditions for a geometric distribution

A random variable follows a geometric distribution if all of these are true:

• There are repeated trials.
• Each trial has only two outcomes: success or failure.
• The probability of success is the same on every trial.
• The trials are independent, meaning one trial does not affect another.
• We are interested in the trial number of the first success.

If these conditions are not true, then the geometric distribution should not be used.

2. Defining the random variable

Let \(X\) be the number of the trial on which the first success occurs.

Then \(X\) can take values \(1, 2, 3, \dots\)

For example:

• \(X = 1\) means the first trial is a success.
• \(X = 2\) means the first trial fails and the second succeeds.
• \(X = 5\) means the first four trials fail and the fifth succeeds.

3. Probability formula

Suppose the probability of success on each trial is \(p\). Then the probability of failure is

\(1-p\).

To get the first success on the \(n\)th trial:

• the first \(n-1\) trials must all be failures, and
• the \(n\)th trial must be a success.

So the formula is

This is the key formula for the geometric distribution.

Why this formula makes sense:

• \((1-p)^{n-1}\) represents \(n-1\) failures in a row,
• then multiplying by \(p\) gives a success on the \(n\)th trial.

4. Probability distribution table pattern

If \(X\) is geometric with success probability \(p\), then:

• \(P(X=1)=p\)
• \(P(X=2)=(1-p)p\)
• \(P(X=3)=(1-p)^2p\)
• \(P(X=4)=(1-p)^3p\)

The probabilities get smaller as \(n\) gets larger, because it becomes less likely that success is delayed for many trials.

5. Worked Example 1: Tossing a coin

A fair coin is tossed repeatedly. Let success mean getting heads. Find the probability that the first heads occurs on the 3rd toss.

Step 1: Identify \(p\)

For a fair coin, \(p = \frac{1}{2}\).

Step 2: Use the formula

Answer: The probability that the first heads occurs on the 3rd toss is \(\frac{1}{8}\).

Check by listing outcomes: The first two tosses must be tails, then the third must be heads: \(TTH\). The probability is

6. Worked Example 2: Multiple-choice guessing

A student guesses on a multiple-choice question with 4 options, only 1 of which is correct. The student keeps guessing on similar independent questions. Find the probability that the first correct answer occurs on the 2nd question.

Step 1: Identify \(p\)

The chance of guessing correctly is \(p=\frac{1}{4}\).

Step 2: Apply the formula

Answer: The probability that the first correct answer occurs on the 2nd question is \(\frac{3}{16}\).

Interpretation: The student must get the first question wrong and the second one right.

7. Worked Example 3: First success on a later trial

A basketball player makes a free throw with probability \(0.7\) on each shot. Assume the shots are independent. Find the probability that the player makes the first successful free throw on the 5th shot.

Step 1: Identify values

• \(p=0.7\)
• \(1-p=0.3\)
• \(n=5\)

Step 2: Use the geometric formula

Answer: The probability is \(0.00567\), or about \(0.567\%\).

This is very small because a player with a high success rate is unlikely to miss the first four shots in a row.

8. Cumulative probability: first success by a certain trial

Sometimes the question is not “on exactly the \(n\)th trial,” but instead “by the \(n\)th trial.”

For example, “What is the probability that the first success happens by the 4th trial?” This means:

This can be found by adding:

But there is a faster way.

The event \(X \leq n\) means there has been at least one success in the first \(n\) trials. The opposite event is no success in the first \(n\) trials, which means all \(n\) trials are failures.

So:

This is often easier to use.

9. Worked Example 4: Success by the 4th trial

A game is won with probability \(0.2\) on each independent play. Find the probability that the first win occurs by the 4th play.

Step 1: Identify \(p\)

\(p=0.2\), so \(1-p=0.8\).

Step 2: Use the cumulative formula

Answer: The probability that the first win occurs by the 4th play is \(0.5904\).

10. Mean of a geometric distribution

The expected value or mean of a geometric distribution tells us the average trial number on which the first success occurs.

If \(X\) has a geometric distribution with success probability \(p\), then

This does not mean success must happen exactly on trial \(\frac{1}{p}\). It means that over many repeated experiments, the average value of \(X\) approaches \(\frac{1}{p}\).

Example: If \(p=0.25\), then

So on average, the first success happens on the 4th trial.

11. Common mistakes to avoid

• Confusing “exactly on the \(n\)th trial” with “by the \(n\)th trial.”
  Exactly on the \(n\)th trial uses \(P(X=n)=(1-p)^{n-1}p\). By the \(n\)th trial uses \(P(X\leq n)=1-(1-p)^n\).
• Using the wrong exponent.
  For \(P(X=n)\), the exponent is \(n-1\), because the first \(n-1\) trials must fail.
• Forgetting what success means.
  Always define success clearly before choosing \(p\).
• Using geometric distribution when trials are not independent.
  If the probability changes from trial to trial, the geometric model does not apply.

12. Quick comparison of ideas

• Exactly on trial \(n\): first \(n-1\) fail, then success
  $$P(X=n)=(1-p)^{n-1}p$$
• By trial \(n\): at least one success in the first \(n\) trials
  $$P(X\leq n)=1-(1-p)^n$$
• Mean trial number of first success:
  $$E(X)=\frac{1}{p}$$

13. Final summary

The geometric distribution is used when we repeat independent trials with the same success probability \(p\), and we want to know when the first success happens.

If \(X\) is the trial number of the first success, then the most important formula is

You should also remember the cumulative formula

and the mean

Whenever you see a question about repeated independent trials and the first success, think of the geometric distribution.