Probability Theory

Sample Spaces and Events are the foundation of probability. Before we can calculate how likely something is, we need to know all the possible outcomes of an experiment and decide which of those outcomes match the event we care about.

In probability, a random experiment is an action or process with an outcome that cannot be predicted with certainty ahead of time. Examples include tossing a coin, rolling a die, or choosing a card from a deck.

This lesson will show you how to describe a random experiment using a sample space, how to define different kinds of events, and how these ideas help us organize probability questions clearly and correctly.

1. What is a sample space?

A sample space is the set of all possible outcomes of a random experiment. We usually use the letter \(S\) to represent the sample space.

If one fair coin is tossed, the possible outcomes are Head and Tail. So the sample space is

$$S = \{H, T\}$$

If a standard six-sided die is rolled once, the sample space is

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

Every valid outcome must appear in the sample space, and nothing impossible should be included. A complete sample space is very important because probabilities are based on it.

2. How to list a sample space

When building a sample space, ask yourself:

• What experiment is happening?
• What are all the possible outcomes?
• How can I represent them clearly?

There are different ways to list sample spaces:

• Listing outcomes directly, such as \(\{1,2,3,4,5,6\}\)
• Using ordered pairs, such as \((1,2)\) for rolling two dice
• Using tables or tree diagrams when there are several stages

For example, if two coins are tossed, the order matters because the first and second toss can be different. The sample space is

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

Notice that \(HT\) and \(TH\) are different outcomes because they happen in a different order.

3. What is an event?

An event is any set of outcomes from the sample space. In other words, an event is a collection of outcomes we are interested in.

If a die is rolled and we want the event “an even number,” then the event is

$$E = \{2,4,6\}$$

Because events are made from outcomes in the sample space, every event is a subset of the sample space.

For the die example, since

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

the event \(E = \{2,4,6\}\) is part of that sample space.

4. Simple events and compound events

A simple event contains exactly one outcome.

For example, when rolling a die, the event “rolling a 3” is

$$A = \{3\}$$

This is a simple event because it has only one outcome.

A compound event contains more than one outcome.

For example, the event “rolling a number greater than 4” is

$$B = \{5,6\}$$

This is a compound event because it has two outcomes.

So:

• Simple event: one outcome
• Compound event: two or more outcomes

5. Complementary events

The complement of an event is everything in the sample space that is not in the event.

If event \(A\) is “rolling an even number” on a die, then

$$A = \{2,4,6\}$$

The complement of \(A\), written as \(A'\) or sometimes \(A^c\), is

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

This means “not rolling an even number.”

Event \(A\) and its complement together make up the whole sample space, and they do not overlap.

In probability, this leads to the useful rule

$$P(A') = 1 - P(A)$$

This rule is helpful when it is easier to find the probability of an event first and then subtract from 1.

6. The empty event and the certain event

Two special events appear often in probability:

• The empty event, written \(\varnothing\), has no outcomes. It is impossible.
• The certain event is the whole sample space \(S\). It must happen.

For example, if a die is rolled:

• “Rolling a 7” is impossible, so the event is \(\varnothing\).
• “Rolling a number less than 7” is certain, so the event is \(S\).

7. Counting outcomes carefully

Many mistakes in probability happen because the sample space is not written carefully. You must make sure each outcome is:

• Possible
• Counted once
• Clearly distinguished from other outcomes

For example, when rolling two dice, the outcome \((2,5)\) is different from \((5,2)\) because the first die and second die are separate.

The sample space for rolling two dice has \(6 \times 6 = 36\) outcomes.

8. Worked Example 1: One coin toss

Experiment: Toss one coin.

Step 1: Write the sample space.

$$S = \{H, T\}$$

Step 2: Define an event. Let event \(A\) be “getting a head.”

$$A = \{H\}$$

Step 3: Classify the event.

Since \(A\) has only one outcome, it is a simple event.

Step 4: Find the complement.

The complement is “not getting a head,” so

$$A' = \{T\}$$

9. Worked Example 2: One die roll

Experiment: Roll a fair six-sided die.

Step 1: Write the sample space.

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

Step 2: Define some events.

• Event \(A\): rolling a prime number
• Event \(B\): rolling a number less than 4

The prime numbers on a die are \(2, 3, 5\), so

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

The numbers less than 4 are \(1,2,3\), so

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

Step 3: Classify the events.

• \(A\) is a compound event because it has three outcomes.
• \(B\) is also a compound event because it has three outcomes.

Step 4: Find a complement.

The complement of \(B\) is “not less than 4,” so

$$B' = \{4,5,6\}$$

10. Worked Example 3: Two coin tosses

Experiment: Toss two coins.

Step 1: Write the sample space.

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

Step 2: Define the event \(C\): getting exactly one head.

Exactly one head happens in outcomes \(HT\) and \(TH\), so

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

Step 3: Is this simple or compound?

Because there are two outcomes, \(C\) is a compound event.

Step 4: Find the complement.

The complement means “not exactly one head,” so the remaining outcomes are \(HH\) and \(TT\).

$$C' = \{HH, TT\}$$

11. Worked Example 4: Rolling two dice

Experiment: Roll two fair six-sided dice.

Step 1: Describe the sample space.

Each outcome is an ordered pair \((x,y)\), where \(x\) is the first die and \(y\) is the second die. Since each die has 6 possible outcomes, there are

$$6 \times 6 = 36$$

possible outcomes in the sample space.

We do not always need to list all 36 outcomes, but they range from \((1,1)\) to \((6,6)\).

Step 2: Define event \(D\): the sum is 7.

The pairs that add to 7 are

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

Step 3: Classify the event.

Since \(D\) has 6 outcomes, it is a compound event.

Step 4: Define event \(E\): doubles.

Doubles means both dice show the same number:

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

This is also a compound event.

12. Connecting sample spaces to probability

Once the sample space and event are written correctly, probability becomes much easier.

If all outcomes in the sample space are equally likely, then the probability of an event \(A\) is

$$P(A) = \frac{\text{number of outcomes in } A}{\text{number of outcomes in } S}$$

For example, if a die is rolled and \(A = \{2,4,6\}\), then

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

This works because there are 3 favorable outcomes out of 6 total outcomes.

13. Common mistakes to avoid

• Leaving out outcomes from the sample space
• Including impossible outcomes
• Ignoring order when order matters, such as with two coin tosses or two dice
• Confusing an event with a single outcome
• Forgetting that the complement includes everything not in the event

Always check: Is my sample space complete? Does my event include exactly the outcomes described in the question?

14. Key ideas to remember

• A sample space is the set of all possible outcomes.
• An event is a set of outcomes from the sample space.
• A simple event has one outcome.
• A compound event has more than one outcome.
• The complement of an event is everything in the sample space that is not in the event.
• Correct probability starts with a correct sample space.

Brief Summary

To solve probability problems, first identify the random experiment and write its sample space. Then describe the event as the set of outcomes you want. Once you can distinguish simple, compound, and complementary events, you can organize probability questions clearly and calculate probabilities with confidence.

Lesson: Theoretical vs. Empirical Probability

Probability is a way of measuring how likely something is to happen. In everyday life, we use probability when we talk about weather forecasts, games, sports, and even medical testing.

In this lesson, you will learn about two important ways to describe probability: theoretical probability and empirical probability. These ideas are closely related, but they are not the same. Understanding the difference helps you decide whether a probability comes from mathematical reasoning or from actual data.

1. What is theoretical probability?

Theoretical probability is the probability you calculate using what you know about all possible outcomes, before doing an experiment many times. It is based on reasoning about equally likely outcomes.

The basic formula is:

$$\text{Theoretical Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$

This works best when the outcomes are equally likely. For example, when flipping a fair coin, the two possible outcomes are heads and tails, and each is equally likely.

So the theoretical probability of getting heads is:

$$P(\text{heads}) = \frac{1}{2}$$

Similarly, for a fair six-sided die, the theoretical probability of rolling a 4 is:

$$P(4) = \frac{1}{6}$$

2. What is empirical probability?

Empirical probability is based on observed results from an experiment or collected data. It tells you how often an event actually happened in a set of trials.

The formula is:

$$\text{Empirical Probability} = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$$

For example, suppose you flip a coin 20 times and get 13 heads. Then the empirical probability of heads is:

$$P(\text{heads}) = \frac{13}{20} = 0.65$$

This result is different from the theoretical probability of \(\frac{1}{2}\). That does not necessarily mean anything is wrong. In a small number of trials, results often vary from the theoretical probability.

3. The key difference

The main difference is where the probability comes from:

• Theoretical probability comes from the structure of the situation and mathematical reasoning.
• Empirical probability comes from experimental data or observations.

Another way to think about it is:

• Theoretical probability asks: What should happen?
• Empirical probability asks: What did happen?

4. Why are they often different?

Theoretical and empirical probabilities are often close, but they are not always exactly the same.

This is because real experiments involve a limited number of trials. Random variation can cause short-run results to differ from expected values.

For example, even though a fair coin has a theoretical probability of \(\frac{1}{2}\) for heads, you might get 7 heads in 10 flips:

$$\frac{7}{10} = 0.7$$

That does not mean the coin must be unfair. It may just be natural variation in a small sample.

As the number of trials increases, the empirical probability usually gets closer to the theoretical probability. This idea is called long-run relative frequency.

So if you flip a fair coin 10 times, the empirical probability might be far from \(0.5\). But if you flip it 10,000 times, the empirical probability will usually be much closer to \(0.5\).

5. When to use each type

Use theoretical probability when:

• All possible outcomes are known.
• The outcomes are equally likely.
• You want to predict probability before collecting data.

Use empirical probability when:

• You have data from experiments or observations.
• The situation may be too complicated to analyze theoretically.
• You want to estimate probability from real-world results.

6. Worked Example 1: Rolling a die

A fair six-sided die is rolled once. Find the theoretical probability of rolling an even number.

Step 1: List the possible outcomes.

The outcomes are \(1,2,3,4,5,6\), so there are 6 total outcomes.

Step 2: Find the favorable outcomes.

The even numbers are \(2,4,6\), so there are 3 favorable outcomes.

Step 3: Use the formula.

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

Answer: The theoretical probability is \(\frac{1}{2}\).

7. Worked Example 2: Probability from data

A student rolls a die 50 times and gets a 6 on 11 rolls. Find the empirical probability of rolling a 6.

Step 1: Count how many times the event happened.

The event “roll a 6” happened 11 times.

Step 2: Count the total number of trials.

The die was rolled 50 times.

Step 3: Use the empirical formula.

$$P(\text{6}) = \frac{11}{50} = 0.22$$

Answer: The empirical probability is \(\frac{11}{50}\), or \(0.22\).

Compare: The theoretical probability of rolling a 6 on a fair die is:

$$\frac{1}{6} \approx 0.167$$

The empirical probability \(0.22\) is higher, but that can happen in a sample of only 50 rolls.

8. Worked Example 3: Comparing theoretical and empirical probability

A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. One marble is selected at random.

Part A: Find the theoretical probability of selecting a blue marble.

Step 1: Find the total number of marbles.

$$5 + 3 + 2 = 10$$

Step 2: Find the number of favorable outcomes.

There are 3 blue marbles.

Step 3: Calculate theoretical probability.

$$P(\text{blue}) = \frac{3}{10} = 0.3$$

Part B: Suppose the experiment is repeated 40 times, and a blue marble is selected 14 times. Find the empirical probability.

$$P(\text{blue}) = \frac{14}{40} = \frac{7}{20} = 0.35$$

Compare:

• Theoretical probability: \(0.3\)
• Empirical probability: \(0.35\)

The two values are close, but not exactly equal. This is normal in experimental data.

9. Worked Example 4: A more realistic situation

A basketball player attempts 80 free throws and makes 62 of them.

Part A: What is the empirical probability that the player makes a free throw?

$$P(\text{make}) = \frac{62}{80} = \frac{31}{40} = 0.775$$

So the empirical probability is \(0.775\), or \(77.5\%\).

Part B: Why is empirical probability more useful than theoretical probability here?

In this situation, there is no simple set of equally likely outcomes like there is with a coin or die. A free throw depends on skill, practice, fatigue, and other real-life factors.

That means empirical probability is the better choice because it uses actual performance data.

10. Important ideas to remember

• Theoretical probability is based on possible outcomes and mathematical reasoning.
• Empirical probability is based on observed results from trials or experiments.
• Empirical probability can change when more data is collected.
• With many trials, empirical probability often gets closer to theoretical probability.
• If outcomes are not equally likely or the situation is based on real-world performance, empirical probability is often more useful.

11. Common mistakes

• Mixing up formulas: theoretical uses favorable outcomes over total possible outcomes, while empirical uses number of times the event happened over total trials.
• Assuming experimental results must exactly match theory: they usually do not, especially with a small number of trials.
• Using theoretical probability when outcomes are not equally likely: not every real-life situation can be handled with simple counting.
• Ignoring sample size: larger samples usually give more reliable empirical estimates.

12. Quick comparison table

• Theoretical probability: based on what should happen
• Empirical probability: based on what actually happened
• Theoretical formula: \(\frac{\text{favorable outcomes}}{\text{total possible outcomes}}\)
• Empirical formula: \(\frac{\text{times event occurred}}{\text{total trials}}\)
• Theoretical example: probability of rolling a 2 on a fair die is \(\frac{1}{6}\)
• Empirical example: if a 2 appears 9 times in 40 rolls, the empirical probability is \(\frac{9}{40}\)

Brief Summary

Theoretical and empirical probability are two different ways to describe how likely an event is. Theoretical probability is found by reasoning about all possible outcomes, while empirical probability is found from experimental data. In small samples, they may differ, but as more trials are performed, empirical probability often moves closer to theoretical probability.

Mutually Exclusive Events and the Addition Rule

In probability, we often want to find the chance that one event or another event happens. For example, what is the probability of rolling a 2 or a 5? Or choosing a card that is a heart or a king?

To answer questions like these, we use the addition rule. This rule helps us combine probabilities correctly, especially when events may overlap.

This lesson will help you:

• understand what mutually exclusive events are,
• tell the difference between events that can happen together and those that cannot,
• use the general addition rule,
• and solve probability problems involving the word or.

1. What does “mutually exclusive” mean?

Two events are mutually exclusive if they cannot happen at the same time.

In other words, if event \(A\) happens, then event \(B\) cannot happen on that same trial. Their overlap is empty.

Mathematically, for mutually exclusive events:

$$P(A \text{ and } B) = 0$$

This means there is no outcome that belongs to both events.

Examples of mutually exclusive events:

• Rolling a die and getting a 3 or a 6 on the same roll
• Choosing one card and getting a queen or a jack
• Tossing one coin and getting heads or tails

Examples that are not mutually exclusive:

• Choosing one card and getting a heart or a king
• Choosing a student who is in soccer or in music, because a student could be in both
• Choosing a number from 1 to 10 that is even or greater than 6, because some numbers satisfy both

2. The meaning of “or” in probability

In probability, the word or usually means at least one of the events happens.

So \(A \text{ or } B\) includes:

• outcomes in \(A\) only,
• outcomes in \(B\) only,
• and outcomes in both \(A\) and \(B\).

This is why we must be careful when adding probabilities. If some outcomes are counted in both events, we could accidentally count them twice.

3. The general addition rule

For any two events \(A\) and \(B\), the probability that \(A\) or \(B\) happens is:

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

This is called the general addition rule.

Why do we subtract \(P(A \text{ and } B)\)? Because if we add \(P(A)\) and \(P(B)\), any outcomes in both events get counted twice. Subtracting the overlap fixes that.

4. Special case: mutually exclusive events

If \(A\) and \(B\) are mutually exclusive, then:

$$P(A \text{ and } B) = 0$$

So the addition rule becomes simpler:

$$P(A \text{ or } B) = P(A) + P(B)$$

This only works when the events cannot happen together.

5. How to decide which rule to use

When you see a probability question involving or, ask yourself:

1. Can the two events happen at the same time?
2. If no, they are mutually exclusive, so add directly.
3. If yes, use the general addition rule and subtract the overlap.

A quick check is to think about one single outcome. Could that outcome belong to both events? If yes, the events are not mutually exclusive.

6. Worked Example 1: Mutually exclusive events on a die

Question: A fair six-sided die is rolled once. What is the probability of getting a 2 or a 5?

Step 1: Define the events.

• \(A\): rolling a 2
• \(B\): rolling a 5

Step 2: Decide if they are mutually exclusive.

On one roll, you cannot get both 2 and 5. So they are mutually exclusive.

Step 3: Find each probability.

Since the die is fair:

$$P(A) = \frac{1}{6}, \quad P(B) = \frac{1}{6}$$

Step 4: Add the probabilities.

$$P(A \text{ or } B) = P(A) + P(B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$

Answer: The probability is \(\frac{1}{3}\).

7. Worked Example 2: Not mutually exclusive events with cards

Question: One card is chosen from a standard 52-card deck. What is the probability that the card is a heart or a king?

Step 1: Define the events.

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

Step 2: Decide if they are mutually exclusive.

No. The king of hearts is both a heart and a king, so the events overlap.

Step 3: Find the probabilities.

• There are 13 hearts, so \(P(A) = \frac{13}{52}\)
• There are 4 kings, so \(P(B) = \frac{4}{52}\)
• There is 1 king of hearts, so \(P(A \text{ and } B) = \frac{1}{52}\)

Step 4: Use the general addition rule.

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$ $$P(A \text{ or } B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$

Answer: The probability is \(\frac{4}{13}\).

Important note: If you had simply added \(\frac{13}{52} + \frac{4}{52}\), you would have counted the king of hearts twice. That is why subtracting the overlap matters.

8. Worked Example 3: Using sets of numbers

Question: A number is chosen at random from 1 to 10. What is the probability that it is even or greater than 6?

Step 1: List the sample space.

The possible outcomes are:

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

Step 2: Define the events.

• \(A\): even numbers = \(\{2,4,6,8,10\}\)
• \(B\): numbers greater than 6 = \(\{7,8,9,10\}\)

Step 3: Find the overlap.

The numbers in both sets are:

$$A \cap B = \{8,10\}$$

So the events are not mutually exclusive.

Step 4: Compute the probabilities.

$$P(A) = \frac{5}{10}, \quad P(B) = \frac{4}{10}, \quad P(A \text{ and } B) = \frac{2}{10}$$

Step 5: Apply the general addition rule.

$$P(A \text{ or } B) = \frac{5}{10} + \frac{4}{10} - \frac{2}{10} = \frac{7}{10}$$

Answer: The probability is \(\frac{7}{10}\).

You can also check by listing all numbers that are even or greater than 6:

$$\{2,4,6,7,8,9,10\}$$

There are 7 favorable outcomes out of 10 total, which confirms \(\frac{7}{10}\).

9. Worked Example 4: Using given probabilities

Question: In a class,

• \(P(S) = 0.55\) is the probability a student plays a sport,
• \(P(M) = 0.40\) is the probability a student studies music,
• \(P(S \text{ and } M) = 0.18\) is the probability a student does both.

Find the probability that a randomly chosen student plays a sport or studies music.

Step 1: Use the general addition rule.

$$P(S \text{ or } M) = P(S) + P(M) - P(S \text{ and } M)$$

Step 2: Substitute the values.

$$P(S \text{ or } M) = 0.55 + 0.40 - 0.18$$ $$P(S \text{ or } M) = 0.77$$

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. Common mistakes to avoid

• Adding probabilities when events overlap.
  Always check whether the events can happen together.
• Confusing “or” with “and”.
  “Or” means at least one happens. “And” means both happen together.
• Forgetting to subtract the overlap.
  If the events are not mutually exclusive, subtract \(P(A \text{ and } B)\).
• Assuming all events with different names are mutually exclusive.
  For example, “heart” and “king” are different descriptions, but one card can be both.

11. A helpful visual idea

Imagine two overlapping circles, one for event \(A\) and one for event \(B\). The overlap represents outcomes that are in both events.

When you calculate \(P(A) + P(B)\), the overlap is counted twice. So you subtract it once:

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

If the circles do not overlap at all, then the events are mutually exclusive, and there is nothing to subtract.

12. Quick guide

• If events cannot happen together:
  $$P(A \text{ or } B) = P(A) + P(B)$$
• If events can happen together:
  $$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$
• Mutually exclusive means:
  $$P(A \text{ and } B) = 0$$

13. Summary

Mutually exclusive events are events that cannot happen at the same time. For these events, the probability of \(A\) or \(B\) is just the sum of their probabilities.

If two events are not mutually exclusive, they overlap. In that case, use the general addition rule:

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

Whenever you see the word or, first ask: Can both events happen together? That question tells you which version of the addition rule to use.

Independent Events and the Multiplication Rule

Probability helps us measure how likely something is to happen. In many problems, we want to know the chance that two events both happen. For example, what is the probability of flipping a head and then rolling an even number? Questions like this are answered using the multiplication rule.

To use this rule correctly, we must understand whether the events are independent or not. This lesson explains what independent events are, how to recognize them, and how to calculate joint probabilities using multiplication.

1. Review: What is an event?

An event is an outcome or a group of outcomes from a probability experiment.

• Rolling a 3 on a die
• Getting heads on a coin toss
• Drawing a red card from a deck

The probability of an event \(A\) is written as \(P(A)\).

All probabilities are between 0 and 1:

$$0 \le P(A) \le 1$$

2. What does “independent” mean?

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

In simple words, knowing that one event happened gives you no new information about whether the other event will happen.

For example:

• Event \(A\): getting heads on a coin toss
• Event \(B\): rolling a 4 on a die

The result of the coin toss does not affect the die roll, so these events are independent.

3. Multiplication rule for independent events

If two events \(A\) and \(B\) are independent, then the probability that both occur is:

$$P(A \text{ and } B) = P(A \cap B) = P(A) \cdot P(B)$$

This is called the multiplication rule for independent events.

The symbol \(A \cap B\) means “\(A\) and \(B\)” or “the intersection of \(A\) and \(B\).”

4. Formal definition of independence

Mathematically, two events \(A\) and \(B\) are independent if and only if:

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

This is not just a rule for calculating. It is also the definition of independence.

So if you are given probabilities and want to prove that two events are independent, you check whether this equation is true.

5. Important idea: independent does not mean mutually exclusive

Students often confuse independent with mutually exclusive.

• Independent events: one event does not affect the other.
• Mutually exclusive events: both events cannot happen at the same time.

For example, when rolling one die:

• Event \(A\): rolling an even number
• Event \(B\): rolling an odd number

These are mutually exclusive because a roll cannot be both even and odd.

But they are not independent, because if you know the roll is even, then the probability of odd becomes 0.

In fact, if two events are mutually exclusive and both have nonzero probability, they cannot be independent.

6. How to identify independent events

Events are often independent in situations like these:

• Tossing a coin and rolling a die
• Two separate spins of a spinner
• Drawing a card, replacing it, then drawing again

Events are often not independent when one result changes the next one, such as drawing cards without replacement.

7. Worked Example 1: Basic multiplication with independent events

A fair coin is tossed and a fair six-sided die is rolled. What is the probability of getting heads and a 5?

Step 1: Find each probability.

$$P(\text{heads}) = \frac{1}{2}$$ $$P(5) = \frac{1}{6}$$

Step 2: Decide if the events are independent.

The coin toss does not affect the die roll, so the events are independent.

Step 3: Multiply.

$$P(\text{heads and }5) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$$

Answer: The probability is \(\frac{1}{12}\).

8. Worked Example 2: More than two independent events

A coin is tossed three times. What is the probability of getting three heads?

Each toss is independent of the others.

The probability of heads on one toss is:

$$P(H) = \frac{1}{2}$$

So the probability of heads on all three tosses is:

$$P(HHH) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$$

Answer: The probability of three heads is \(\frac{1}{8}\).

This shows that for several independent events, we multiply all the individual probabilities.

9. Worked Example 3: Testing whether events are independent

Suppose:

$$P(A) = 0.4, \quad P(B) = 0.5, \quad P(A \cap B) = 0.2$$

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

Step 1: Use the definition of independence.

If the events are independent, then:

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

Step 2: Calculate the product.

$$P(A)P(B) = 0.4 \times 0.5 = 0.2$$

Step 3: Compare.

$$P(A \cap B) = 0.2$$

Since:

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

the events are independent.

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

10. Worked Example 4: A case that is not independent

A bag contains 3 red marbles and 2 blue marbles. Two marbles are drawn without replacement. What is the probability that both are red?

Step 1: Find the probability of the first red marble.

$$P(\text{first red}) = \frac{3}{5}$$

Step 2: Find the probability of the second red marble after the first red was drawn.

Now there are 2 red marbles left out of 4 marbles total.

$$P(\text{second red} \mid \text{first red}) = \frac{2}{4} = \frac{1}{2}$$

Step 3: Multiply the probabilities.

$$P(\text{both red}) = \frac{3}{5} \cdot \frac{1}{2} = \frac{3}{10}$$

Answer: The probability is \(\frac{3}{10}\).

Why are these events not independent?

Before any draw, the probability of red is \(\frac{3}{5}\). After drawing one red marble, the probability of red on the second draw becomes \(\frac{2}{4}\). Since the probability changed, the events are not independent.

11. Connecting independence to conditional probability

Another way to think about independence is with conditional probability.

If events \(A\) and \(B\) are independent, then:

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

This means the probability of \(A\) given that \(B\) happened is the same as the original probability of \(A\).

Similarly:

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

This matches the idea that one event does not affect the other.

12. Steps for solving independence problems

1. Identify the two events clearly.
2. Decide whether the events are independent.
3. If they are independent, use: $$P(A \cap B) = P(A)P(B)$$
4. If you are testing for independence, compare \(P(A \cap B)\) with \(P(A)P(B)\).
5. Be careful with situations where outcomes change, such as drawing without replacement.

13. Common mistakes to avoid

• Adding instead of multiplying when asked for the probability that both events happen.
• Assuming events are independent without checking the situation carefully.
• Confusing independent with mutually exclusive.
• Forgetting that “without replacement” usually means the events are dependent.

14. Quick practice ideas

Ask yourself these questions when solving:

• Does the first event change the second event?
• Am I looking for “and,” meaning both events happen?
• Should I multiply the probabilities?
• Can I test independence using \(P(A \cap B) = P(A)P(B)\)?

15. Summary

Independent events are events where one does not affect the probability of the other. For independent events, the probability that both occur is found by multiplying their probabilities:

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

This formula is also the definition of independence. To prove events are independent, check whether the probability of both events equals the product of their individual probabilities.

When solving problems, always pay attention to whether the first outcome changes the second. If it does, the events are not independent, and you should not use the independent multiplication rule directly.

Conditional Probability is used when we want the probability of an event given that another event has already happened.

For example, suppose you draw one card from a standard deck. The probability of drawing a king is usually based on all 52 cards. But if you are told the card is a face card, then you no longer consider all 52 cards. You only consider the face cards. This smaller set of possible outcomes is called a reduced sample space.

That is the main idea of conditional probability: when new information is given, the total number of possible outcomes may change.

The notation for conditional probability is \(P(A\mid B)\), which is read as “the probability of event \(A\) given event \(B\).”

The formula is

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

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

Here:

• \(A\cap B\) means both \(A\) and \(B\) happen.
• \(P(B)\) is the probability that the given condition happens.
• \(P(A\mid B)\) is the probability of \(A\) within the outcomes where \(B\) has happened.

Why does this formula make sense?

Once we know \(B\) has happened, we ignore all outcomes not in \(B\). So the “new total” is \(P(B)\). Out of those outcomes, we want the part where both \(A\) and \(B\) happen, which is \(P(A\cap B)\).

So conditional probability is really a fraction:

$$\text{conditional probability} = \frac{\text{part you want inside the condition}}{\text{whole conditioned group}}$$

Important idea: conditional probability is not always the same as ordinary probability.

For instance, the probability that a student is left-handed in a school might be different from the probability that a student is left-handed given that the student plays baseball. The extra information can change the probability.

Visual way to think about it

Imagine all outcomes in a rectangle. Event \(B\) is one region inside it. Event \(A\) is another region. The overlap \(A\cap B\) is the part where both occur. When finding \(P(A\mid B)\), you focus only on the region for \(B\), and then ask what fraction of that region lies in the overlap.

Worked Example 1: Rolling a die

A fair six-sided die is rolled.

• Let \(A\) = “the result is even.”
• Let \(B\) = “the result is greater than 3.”

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

Step 1: List the outcomes in \(B\).

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

Step 2: Find which of these are also in \(A\).

The even numbers in \(\{4,5,6\}\) are \(\{4,6\}\).

Step 3: Form the conditional probability.

There are 3 outcomes in the reduced sample space, and 2 of them satisfy \(A\).

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

Check using the formula:

\(A\cap B\) means even and greater than 3, so the outcomes are \(\{4,6\}\). Thus \(P(A\cap B)=\frac{2}{6}=\frac{1}{3}\).

Also, \(P(B)=\frac{3}{6}=\frac{1}{2}\).

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

Both methods agree.

Worked Example 2: Cards

One card is drawn from a standard 52-card deck.

• Let \(A\) = “the card is a king.”
• Let \(B\) = “the card is a face card.”

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

Step 1: Identify the conditioned group.

Face cards are jacks, queens, and kings. There are 3 face cards in each of 4 suits, so there are 12 face cards total.

Step 2: Count how many of those are kings.

There are 4 kings in the deck, and all of them are face cards.

Step 3: Compute the probability in the reduced sample space.

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

Using the formula:

\(P(A\cap B)=P(\text{king and face card})=P(\text{king})=\frac{4}{52}\).

Also, \(P(B)=\frac{12}{52}\).

$$P(A\mid B)=\frac{4/52}{12/52}=\frac{4}{12}=\frac{1}{3}$$

This shows how knowing the card is a face card changes the sample space from 52 cards to 12 cards.

Worked Example 3: Table of data

A class survey gives the following information:

	Plays Sport	Does Not Play Sport	Total
Male	18	7	25
Female	12	13	25
Total	30	20	50

Find the probability that a student is male given that the student plays a sport.

Let

• \(A\) = “student is male”
• \(B\) = “student plays a sport”

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

Method 1: Read from the reduced sample space.

If we already know the student plays a sport, then we only look at the “Plays Sport” column. There are 30 such students.

Among those 30 students, 18 are male.

$$P(A\mid B)=\frac{18}{30}=\frac{3}{5}=0.6$$

Method 2: Use the formula.

\(P(A\cap B)=P(\text{male and plays sport})=\frac{18}{50}\).

\(P(B)=P(\text{plays sport})=\frac{30}{50}\).

$$P(A\mid B)=\frac{18/50}{30/50}=\frac{18}{30}=\frac{3}{5}$$

So the probability is \(0.6\), or 60%.

Worked Example 4: Two marbles without replacement

A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn one after the other without replacement.

Find the probability that the second marble is red given that the first marble was blue.

Let

• \(A\) = “second marble is red”
• \(B\) = “first marble is blue”

If the first marble was blue, then one blue marble has been removed. The bag now has 5 red marbles and 2 blue marbles, for a total of 7 marbles.

So, in this reduced sample space, the probability that the second marble is red is

$$P(A\mid B)=\frac{5}{7}$$

Why conditional probability matters here: the first draw changes the situation. Once we know the first marble was blue, the probabilities for the second draw are based on the remaining marbles, not the original 8 marbles.

Key points to remember

• Conditional probability means a probability with extra information.
• The condition changes the sample space.
• The formula is $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$ provided \(P(B)\neq 0\).
• Always ask: “Given \(B\), what outcomes are still possible?”
• Then find how many of those outcomes also satisfy \(A\).

Common mistakes

• Using the original sample space instead of the reduced one. If a condition is given, you usually should not use the original total.
• Mixing up \(P(A\mid B)\) and \(P(B\mid A)\). These are usually different. “A given B” is not the same as “B given A.”
• Forgetting the intersection. In the formula, the numerator is \(P(A\cap B)\), the probability that both happen.
• Dividing by the wrong probability. The denominator is the probability of the condition, which is \(P(B)\).

Quick comparison

• \(P(A)\): probability of \(A\) out of all outcomes.
• \(P(A\mid B)\): probability of \(A\) out of only the outcomes where \(B\) happens.

Mini-check questions

1. If a card is known to be a heart, what sample space should you use: all 52 cards or just the hearts?
2. In \(P(A\mid B)\), which event goes in the denominator?
3. If \(A\cap B\) is impossible, what is \(P(A\mid B)\)?

Answers

1. Just the hearts, because the condition reduces the sample space.
2. \(P(B)\), because \(B\) is the given condition.
3. It is 0, because \(P(A\cap B)=0\).

Summary

Conditional probability tells us how likely an event is when another event is already known to have happened. It works by reducing the sample space to only the outcomes that satisfy the condition.

The main formula is

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

To solve problems, first identify the condition, then focus only on that smaller group of outcomes. Finally, count how many of those outcomes also satisfy the event you want.

Tree Diagrams and Sequential Probability

In probability, many situations happen in steps. For example, you might toss a coin twice, pick two cards one after another, or choose students from a class without replacing them. When an event happens in stages, we call it a sequential probability problem.

A very useful way to organize these problems is with a tree diagram. A tree diagram shows each possible outcome step by step, and it helps us calculate the probability of each final result.

In this lesson, you will learn how to build tree diagrams, how to multiply probabilities along branches, and how to add probabilities when different branches lead to the same type of outcome.

1. What is a tree diagram?

A tree diagram is a branching picture that represents all possible outcomes of a sequence of events. Each branch shows one possible result at a stage, and each branch is labeled with its probability.

The diagram starts from one point. From there, branches split for the first event. Then each of those branches splits again for the second event, and so on.

For example, if you flip a coin twice, the first flip can be Heads or Tails. Then from each of those, the second flip can also be Heads or Tails. This creates four possible final outcomes:

• HH
• HT
• TH
• TT

2. Key idea: multiply along a path

When you follow one complete path in a tree diagram, you are finding the probability that all of those events happen in that order. To calculate that, you multiply the probabilities along the path.

If event A happens first and event B happens second, then:

$$P(A \text{ then } B)=P(A)\times P(B \mid A)$$

The symbol \(P(B \mid A)\) means “the probability of B given that A has already happened.” This is especially important when the second event depends on the first one.

3. Key idea: add across different paths

Sometimes there is more than one path that gives the outcome you want. In that case, after finding each path probability, you add them.

So tree-diagram problems often use this pattern:

• Multiply along each branch
• Add the relevant final branches

4. Independent and dependent events

To use a tree diagram correctly, you need to know whether events are independent or dependent.

Independent events are events where one event does not change the probability of the next event.

Examples:

• Flipping a coin twice
• Rolling a die, then rolling it again

If two events are independent, then:

$$P(A \text{ then } B)=P(A)\times P(B)$$

Dependent events are events where the first event changes the probability of the second event.

Examples:

• Drawing two cards from a deck without replacement
• Choosing two students from a group without putting the first one back

In dependent events, the branch probabilities in later stages may change based on what happened earlier.

5. How to build a tree diagram

Follow these steps:

1. Identify each stage of the event.
2. Draw branches for all outcomes of the first stage.
3. From each branch, draw branches for the second stage.
4. Label every branch with its correct probability.
5. Multiply probabilities along a path to find a final outcome.
6. Add path probabilities if more than one path matches the event you want.

Worked Example 1: Tossing a coin twice

A fair coin is tossed twice. Find the probability of getting exactly one head.

Step 1: List the branches

For each toss:

• Heads: \(\frac{1}{2}\)
• Tails: \(\frac{1}{2}\)

The possible outcomes are:

• HH
• HT
• TH
• TT

Step 2: Find the probability of each path

For example:

$$P(HT)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$$ $$P(TH)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$$

Step 3: Add the paths with exactly one head

Exactly one head happens in HT or TH.

$$P(\text{exactly one head})=P(HT)+P(TH)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

Answer: The probability of getting exactly one head is \(\frac{1}{2}\).

Worked Example 2: Choosing colored marbles without replacement

A bag contains 3 red marbles and 2 blue marbles. Two marbles are chosen without replacement. Find the probability that both marbles are red.

Step 1: First draw probabilities

There are 5 marbles total.

$$P(R_1)=\frac{3}{5}, \qquad P(B_1)=\frac{2}{5}$$

Step 2: Second draw probabilities

If the first marble was red, then 2 red marbles remain out of 4 total:

$$P(R_2 \mid R_1)=\frac{2}{4}=\frac{1}{2}$$

If the first marble was blue, then 3 red marbles remain out of 4 total:

$$P(R_2 \mid B_1)=\frac{3}{4}$$

The probabilities changed, so this is a dependent event.

Step 3: Multiply along the RR branch

$$P(RR)=P(R_1)\times P(R_2 \mid R_1)=\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$$

Answer: The probability that both marbles are red is \(\frac{3}{10}\).

Worked Example 3: Finding “one of each color”

Using the same bag with 3 red marbles and 2 blue marbles, two marbles are chosen without replacement. Find the probability of getting one red and one blue.

This outcome can happen in two different orders:

• Red then Blue
• Blue then Red

Path 1: Red then Blue

$$P(RB)=\frac{3}{5}\times\frac{2}{4}=\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$$

Path 2: Blue then Red

$$P(BR)=\frac{2}{5}\times\frac{3}{4}=\frac{6}{20}=\frac{3}{10}$$

Add the two paths

$$P(\text{one red and one blue})=P(RB)+P(BR)=\frac{3}{10}+\frac{3}{10}=\frac{6}{10}=\frac{3}{5}$$

Answer: The probability of choosing one red and one blue is \(\frac{3}{5}\).

Worked Example 4: A three-stage tree diagram

A factory machine produces a good item with probability \(0.9\) and a defective item with probability \(0.1\). Three items are checked, and we assume each item is independent of the others. Find the probability that exactly two of the three items are good.

Step 1: Understand the branches

For each item:

• Good (G): \(0.9\)
• Defective (D): \(0.1\)

Step 2: Identify outcomes with exactly two good items

• GGD
• GDG
• DGG

Step 3: Find each path probability

$$P(GGD)=0.9\times0.9\times0.1=0.081$$ $$P(GDG)=0.9\times0.1\times0.9=0.081$$ $$P(DGG)=0.1\times0.9\times0.9=0.081$$

Step 4: Add the paths

$$P(\text{exactly two good})=0.081+0.081+0.081=0.243$$

Answer: The probability that exactly two of the three items are good is \(0.243\).

6. Reading a tree diagram carefully

When looking at a tree diagram, remember:

• Each level represents a new stage in the experiment.
• The probabilities on branches from the same point should add up to 1.
• The probability at the end of a path comes from multiplying all branch probabilities on that path.
• If an event can happen in several ways, add the probabilities of those final branches.

7. Common mistakes to avoid

• Adding when you should multiply: probabilities along one path must be multiplied.
• Forgetting to add multiple paths: if an outcome can happen in different orders, include all of them.
• Using the same probabilities after no replacement: in dependent events, probabilities change.
• Missing branches: make sure the tree shows all possible outcomes.
• Ignoring order: in sequential probability, red then blue is a different path from blue then red, even if they belong to the same final event.

8. Quick strategy for solving problems

When you see a sequential probability question, ask yourself:

1. How many stages are there?
2. What are the possible outcomes at each stage?
3. Are the events independent or dependent?
4. What final outcome am I looking for?
5. Do I need one path or several paths?

Then build the tree, multiply along the needed paths, and add if necessary.

9. Summary

Tree diagrams help organize probability problems that happen in stages. They are especially useful for showing all possible outcomes clearly.

To solve these problems, multiply probabilities along a branch and add probabilities across different branches that match the event you want.

For independent events, the probabilities stay the same at each stage. For dependent events, later probabilities can change based on earlier outcomes.

If you build the tree carefully and keep track of which branches matter, you can solve many sequential probability problems accurately and confidently.

Combinatorial Probability is a way of finding probability by counting. Instead of listing every possible outcome one by one, we use counting methods such as permutations and combinations to count how many outcomes are possible and how many are favorable.

This is especially useful when there are many outcomes, such as choosing students from a class, arranging letters, or drawing cards from a deck.

The main idea is this:

$$P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{number of total outcomes}}$$

In combinatorial probability, the challenge is usually not the probability formula itself. The challenge is counting correctly.

1. Start with equally likely outcomes

Combinatorial probability works best when all outcomes in the sample space are equally likely. For example, when drawing a random hand of cards, every possible hand of that size is equally likely.

If outcomes are equally likely, then counting methods can be used directly. If outcomes are not equally likely, then simple counting may not give the correct probability.

2. Total outcomes and favorable outcomes

To calculate a combinatorial probability, you usually do two things:

• Count the total number of possible outcomes.
• Count the number of outcomes that satisfy the condition.

Then divide:

$$P(E) = \frac{n(E)}{n(S)}$$

Here, \(n(E)\) means the number of favorable outcomes, and \(n(S)\) means the number of outcomes in the sample space.

3. Permutations: when order matters

A permutation is an arrangement where order matters.

For example, choosing 1st, 2nd, and 3rd place in a race is a permutation, because changing the order changes the result.

The number of permutations of \(r\) objects chosen from \(n\) objects is:

$$^nP_r = \frac{n!}{(n-r)!}$$

The symbol \(!\) means factorial:

$$n! = n(n-1)(n-2)\cdots(2)(1)$$

For example,

$$5! = 5\cdot4\cdot3\cdot2\cdot1 = 120$$

4. Combinations: when order does not matter

A combination is a selection where order does not matter.

For example, choosing 3 students to form a committee is a combination, because the group is the same no matter what order the names are listed in.

The number of combinations of \(r\) objects chosen from \(n\) objects is:

$$^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

5. How to decide: permutation or combination?

• Use a permutation if rearranging the chosen items creates a different outcome.
• Use a combination if rearranging the chosen items does not create a different outcome.

A good question to ask is: Does order matter here?

6. Common situations in combinatorial probability

• Selecting groups: usually combinations.
• Ranking or arranging: usually permutations.
• Card hands: combinations, because a hand is the same regardless of order.
• Sequences of outcomes: often permutations or multiplication counting.

7. Worked Example 1: Choosing students

A class has 10 students. Two students are chosen at random to clean the board. What is the probability that both chosen students are girls if there are 6 girls and 4 boys?

Step 1: Count total outcomes.

We are choosing 2 students from 10, and order does not matter.

$$\text{Total outcomes} = \binom{10}{2} = \frac{10!}{2!8!} = 45$$

Step 2: Count favorable outcomes.

We need both students to be girls. There are 6 girls, and we choose 2 of them.

$$\text{Favorable outcomes} = \binom{6}{2} = \frac{6!}{2!4!} = 15$$

Step 3: Find the probability.

$$P(\text{both girls}) = \frac{15}{45} = \frac{1}{3}$$

Answer: The probability is \(\frac{1}{3}\).

Why combinations? Because choosing Anna and Mia is the same group as choosing Mia and Anna.

8. Worked Example 2: Arranging letters

The letters A, B, C, D, and E are arranged randomly. What is the probability that the arrangement begins with A and ends with E?

Step 1: Count total outcomes.

There are 5 distinct letters. The number of arrangements is:

$$5! = 120$$

Step 2: Count favorable outcomes.

If the arrangement must begin with A and end with E, then those two positions are fixed.

That leaves B, C, and D to arrange in the 3 middle positions.

$$3! = 6$$

Step 3: Find the probability.

$$P(\text{begins with A and ends with E}) = \frac{3!}{5!} = \frac{6}{120} = \frac{1}{20}$$

Answer: The probability is \(\frac{1}{20}\).

Why permutations? Because different orders of the letters are different outcomes.

9. Worked Example 3: Drawing cards

From a standard deck of 52 cards, 5 cards are drawn at random. What is the probability that all 5 cards are hearts?

Step 1: Count total outcomes.

A 5-card hand is a selection, so order does not matter.

$$\text{Total outcomes} = \binom{52}{5}$$

Step 2: Count favorable outcomes.

There are 13 hearts in the deck. To get all 5 cards as hearts, we choose 5 from those 13 hearts.

$$\text{Favorable outcomes} = \binom{13}{5}$$

Step 3: Write the probability.

$$P(\text{all 5 hearts}) = \frac{\binom{13}{5}}{\binom{52}{5}}$$

Now calculate:

$$\binom{13}{5} = 1287$$

$$\binom{52}{5} = 2,598,960$$

So,

$$P(\text{all 5 hearts}) = \frac{1287}{2,598,960} \approx 0.000495$$

Answer: The probability is approximately \(0.000495\), or about \(0.0495\%\).

10. Worked Example 4: At least one

A bag contains 7 red marbles and 5 blue marbles. Three marbles are chosen at random without replacement. What is the probability that at least one marble is blue?

Problems with the phrase at least one are often easier if we use the complement.

Complement idea:

$$P(\text{at least one blue}) = 1 - P(\text{no blue})$$

Step 1: Count total outcomes.

There are 12 marbles total, and we choose 3.

$$\text{Total outcomes} = \binom{12}{3} = 220$$

Step 2: Count the complement.

No blue means all 3 marbles are red. There are 7 red marbles.

$$\text{No blue outcomes} = \binom{7}{3} = 35$$

So,

$$P(\text{no blue}) = \frac{35}{220} = \frac{7}{44}$$

Step 3: Use the complement.

$$P(\text{at least one blue}) = 1 - \frac{7}{44} = \frac{37}{44}$$

Answer: The probability is \(\frac{37}{44}\).

11. Helpful strategies for solving combinatorial probability problems

• Read carefully and decide what counts as one outcome.
• Ask whether order matters.
• Count total outcomes first.
• Then count favorable outcomes.
• For phrases like at least one, at most, or none, consider using the complement.
• Check that favorable outcomes are part of the same sample space as total outcomes.

12. Common mistakes to avoid

• Using permutations when combinations are needed, or the other way around.
• Forgetting that order does not matter in selections like committees or hands of cards.
• Counting favorable outcomes and total outcomes in different ways. For example, using ordered outcomes for one and unordered outcomes for the other.
• Ignoring the complement method when it would make the problem easier.
• Mixing up “without replacement” and “with replacement”. In combinatorial probability, many selection problems are without replacement.

13. Quick comparison

• Permutation: arrangement, ranking, ordered list
• Combination: group, team, hand, committee

14. Final summary

Combinatorial probability uses counting methods to find the probability of an event.

You begin with

$$P(E) = \frac{n(E)}{n(S)}$$

Then you count carefully using:

• Permutations when order matters: $$^nP_r = \frac{n!}{(n-r)!}$$
• Combinations when order does not matter: $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

The key skill is deciding how to count outcomes correctly. Once you know whether order matters, many probability problems become much easier to solve.

Expected Value is a way of finding the long-run average outcome of a random situation.

It helps answer questions like:

• How much money do you expect to win or lose in a game?
• What is the average score you would get over many trials?
• Is a game fair?
• Which financial choice is better when outcomes are uncertain?

Even though the actual result of one trial may be different, the expected value tells you what the average result would approach if the situation were repeated many times.

For example, if a game has an expected value of \(-\$0.50\), that does not mean you lose exactly 50 cents every time. It means that over many plays, your average result would be a loss of about 50 cents per game.

1. Discrete random variables

A random variable is a variable whose value depends on chance.

A discrete random variable takes specific separate values, often whole numbers. For example:

• the number shown on a die: \(1,2,3,4,5,6\)
• the number of heads in 3 coin tosses: \(0,1,2,3\)
• the profit from a game: \(-2, 0, 5\)

To work with expected value, we first describe the random variable using a probability distribution.

A probability distribution lists:

• all possible values of the random variable
• the probability of each value

If \(X\) is a discrete random variable, then a probability distribution may look like this:

$$ \begin{array}{c|cccc} X & x_1 & x_2 & x_3 & x_4 \\ \hline P(X=x) & p_1 & p_2 & p_3 & p_4 \end{array} $$

The probabilities must satisfy two rules:

• each probability is between 0 and 1
• the probabilities add to 1

2. Formula for expected value

The expected value of a discrete random variable \(X\) is written as \(E(X)\).

It is found by multiplying each possible value by its probability, then adding the results:

$$ E(X)=\sum xP(X=x) $$

In expanded form, this means:

$$ E(X)=x_1p_1+x_2p_2+x_3p_3+\cdots $$

This is a weighted average. Values with higher probabilities affect the expected value more.

3. How to calculate expected value

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

4. Worked Example 1: Expected value of a die roll

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

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

$$ E(X)=1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)+5\left(\frac{1}{6}\right)+6\left(\frac{1}{6}\right) $$ $$ E(X)=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=3.5 $$

The expected value is \(3.5\).

This does not mean you can roll a 3.5 on one toss. It means that over many rolls, the average result will approach 3.5.

5. Worked Example 2: A simple game

You pay \(\$2\) to play a game. A fair coin is tossed.

• If it lands heads, you win \(\$5\).
• If it lands tails, you win nothing.

We want the net gain, which means winnings minus cost.

So the possible outcomes for net gain are:

• Heads: \(5-2=3\), so gain \(\$3\)
• Tails: \(0-2=-2\), so gain \(-\$2\)

Let \(X\) be the net gain. Then:

$$ P(X=3)=\frac{1}{2}, \qquad P(X=-2)=\frac{1}{2} $$

Now calculate expected value:

$$ E(X)=3\left(\frac{1}{2}\right)+(-2)\left(\frac{1}{2}\right) $$ $$ E(X)=\frac{3}{2}-\frac{2}{2}=\frac{1}{2}=0.5 $$

The expected value is \(\$0.50\).

This means that in the long run, the average net gain is 50 cents per game. This is a good game for the player.

6. Fair games

A game is called fair when the expected net gain is 0.

That means, on average, neither the player nor the organiser gains money in the long run.

• If \(E(X)=0\), the game is fair.
• If \(E(X)>0\), the game favors the player.
• If \(E(X)<0\), the game favors the organiser.

When a question asks whether a game is fair, make sure you use the net gain, not just the prize amount.

7. Worked Example 3: Is this game fair?

A bag contains 5 balls: 1 red and 4 blue. You pay \(\$3\) to draw one ball.

• If you draw the red ball, you receive \(\$10\).
• If you draw a blue ball, you receive nothing.

Find the expected net gain and decide if the game is fair.

Step 1: Find the possible net gains.

• Red ball: \(10-3=7\), so net gain is \(\$7\)
• Blue ball: \(0-3=-3\), so net gain is \(-\$3\)

Step 2: Find the probabilities.

$$ P(X=7)=\frac{1}{5}, \qquad P(X=-3)=\frac{4}{5} $$

Step 3: Calculate expected value.

$$ E(X)=7\left(\frac{1}{5}\right)+(-3)\left(\frac{4}{5}\right) $$ $$ E(X)=\frac{7}{5}-\frac{12}{5}=-\frac{5}{5}=-1 $$

The expected net gain is \(-\$1\).

This means that over many plays, a player would lose an average of \(\$1\) per game. So the game is not fair; it favors the organiser.

8. Expected value in financial decisions

Expected value is also used to compare choices involving uncertainty.

For example, a business might compare possible profits under different market conditions, or a person might compare whether buying insurance is worth it.

The idea stays the same: multiply each possible outcome by its probability, then add.

Worked Example 4: Choosing between two options

A student can choose one of two weekend jobs.

Job A:

• 50% chance of earning \(\$80\)
• 50% chance of earning \(\$120\)

Job B:

• 20% chance of earning \(\$50\)
• 50% chance of earning \(\$100\)
• 30% chance of earning \(\$150\)

Find the expected earning for each job.

Job A

$$ E(A)=80(0.5)+120(0.5) $$ $$ E(A)=40+60=100 $$

Expected earning for Job A is \(\$100\).

Job B

$$ E(B)=50(0.2)+100(0.5)+150(0.3) $$ $$ E(B)=10+50+45=105 $$

Expected earning for Job B is \(\$105\).

Since \(105>100\), Job B has the higher expected earning.

This does not guarantee Job B will pay more on one particular weekend. It only means its average earning is higher over many weekends.

9. Important ideas and common mistakes

• Expected value is an average in the long run. It is not usually the result of one single trial.
• Use all possible outcomes. Leaving one out gives the wrong answer.
• Check that probabilities add to 1. If they do not, the distribution is incomplete or incorrect.
• Use net gain for games involving a cost. Subtract the cost to play before finding expected value.
• Negative expected value means average loss. Positive expected value means average gain.

10. Quick practice thinking

If a game costs \(\$4\) to play and gives a \(\frac{1}{10}\) chance of winning \(\$30\), with no prize otherwise, then the net gains are:

• win: \(30-4=26\)
• lose: \(0-4=-4\)

The expected value is:

$$ E(X)=26\left(\frac{1}{10}\right)+(-4)\left(\frac{9}{10}\right) $$ $$ E(X)=2.6-3.6=-1 $$

So the player expects to lose \(\$1\) per game on average.

11. Summary

Expected value tells us the long-run average result of a random process.

For a discrete random variable \(X\), the formula is:

$$ E(X)=\sum xP(X=x) $$

To calculate it, multiply each possible value by its probability and add the results.

Expected value is useful for analyzing probability distributions, deciding whether games are fair, and comparing financial choices with uncertain outcomes.

Always remember: expected value describes what happens on average over many trials, not what must happen in one attempt.