Probability Theory and Combinatorics

Sample Spaces and the Law of Large Numbers

Probability is the math of chance. It helps us describe how likely an event is to happen.

To understand probability well, we first need to know all possible outcomes of an experiment. This list of outcomes is called the sample space.

We also need to understand that when we repeat an experiment many times, the results we actually observe often get closer to the probability we expected. This idea is called the Law of Large Numbers.

In this lesson, you will learn how to:

• find the sample space of an experiment,
• use the sample space to calculate probability,
• compare theoretical probability and experimental probability,
• understand the Law of Large Numbers.

1. What is a sample space?

A sample space is the set of all possible outcomes of a probability experiment.

An outcome is one possible result. For example, if you toss a coin once, the outcomes are Heads and Tails.

We can write the sample space using braces:

$$S = \{\text{Heads}, \text{Tails}\}$$

The sample space must include every valid outcome, but it should not include impossible outcomes.

Examples of sample spaces

• Rolling one number cube: $$S = \{1,2,3,4,5,6\}$$
• Choosing a day of the weekend: $$S = \{\text{Saturday}, \text{Sunday}\}$$
• Tossing two coins: $$S = \{HH, HT, TH, TT\}$$

2. Why sample spaces matter

The sample space helps us count how many outcomes are possible. Once we know that, we can calculate probability.

If all outcomes are equally likely, the probability of an event is:

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

A favorable outcome is an outcome that matches the event we want.

For example, when rolling a fair die, the probability of getting an even number is based on the sample space $$\{1,2,3,4,5,6\}$$. The even numbers are $$2,4,6$$, so there are 3 favorable outcomes out of 6 total outcomes.

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

3. Listing sample spaces

For simple experiments, we can list the sample space directly. For more than one action, we must be careful to include every possible combination.

Here are some useful ways to build a sample space:

• List outcomes one by one.
• Use a table for two-step experiments.
• Use a tree diagram to organize outcomes.

Example: Tossing two coins

The first coin can be H or T. The second coin can also be H or T.

So the sample space is:

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

Notice that HT and TH are different outcomes because the order matters. In one outcome, the first coin is Heads and the second is Tails. In the other, the first coin is Tails and the second is Heads.

4. Theoretical probability

Theoretical probability is the probability we calculate using math before doing the experiment.

It is based on the sample space and assumes the experiment is fair.

For example, if we toss two fair coins, what is the probability of getting exactly one head?

The sample space is:

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

The outcomes with exactly one head are $$HT$$ and $$TH$$. That is 2 favorable outcomes out of 4 total outcomes.

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

5. Experimental probability

Experimental probability is based on actual results from performing an experiment.

Its formula is:

$$P(\text{event}) = \frac{\text{number of times the event happened}}{\text{total number of trials}}$$

For example, suppose a coin is tossed 20 times and it lands on Heads 13 times. Then the experimental probability of Heads is:

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

This is not exactly $$\frac{1}{2}$$, even though a fair coin has theoretical probability $$\frac{1}{2}$$ of landing on Heads.

That is normal. In a small number of trials, results can vary quite a bit.

6. The Law of Large Numbers

The Law of Large Numbers says that as an experiment is repeated many times, the experimental probability tends to get closer to the theoretical probability.

This does not mean the results become perfect every time. It means the overall pattern becomes more stable when the number of trials is large.

For a fair coin:

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

If you toss the coin:

• 10 times, you might get 7 heads, so $$\frac{7}{10} = 0.7$$
• 100 times, you might get 53 heads, so $$\frac{53}{100} = 0.53$$
• 1000 times, you might get 497 heads, so $$\frac{497}{1000} = 0.497$$

As the number of tosses increases, the experimental probability often gets closer to $$0.5$$.

Important idea: The Law of Large Numbers does not guarantee short-term results. It describes what happens over many trials.

7. Worked Examples

Example 1: Find a sample space

Question: Write the sample space for rolling one fair die.

Step 1: Think about every possible result.

A die has six faces labeled 1 through 6.

Answer:

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

Example 2: Use a sample space to find probability

Question: If one fair die is rolled, what is the probability of rolling a number greater than 4?

Step 1: Write the sample space.

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

Step 2: Identify favorable outcomes.

Numbers greater than 4 are $$5$$ and $$6$$.

Step 3: Count outcomes.

• Favorable outcomes: 2
• Total outcomes: 6

Step 4: Calculate probability.

$$P(\text{greater than 4}) = \frac{2}{6} = \frac{1}{3}$$

Example 3: Two-step experiment

Question: Two coins are tossed. What is the probability of getting at least one head?

Step 1: Write the sample space.

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

Step 2: Find outcomes with at least one head.

The outcomes are $$HH, HT, TH$$.

Step 3: Count outcomes.

• Favorable outcomes: 3
• Total outcomes: 4

Step 4: Calculate probability.

$$P(\text{at least one head}) = \frac{3}{4}$$

Example 4: Apply the Law of Large Numbers

Question: A bag contains 5 red marbles and 5 blue marbles. One marble is chosen, the color is recorded, and then the marble is put back. What should happen to the experimental probability of choosing red after many trials?

Step 1: Find the theoretical probability.

There are 10 marbles in total, and 5 are red.

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

Step 2: Use the Law of Large Numbers.

If you repeat the experiment many times, the experimental probability of red should get closer to $$\frac{1}{2}$$.

Answer: Over many trials, the fraction of times red is chosen should approach $$0.5$$.

8. Common mistakes to avoid

• Leaving out outcomes from the sample space.
• Counting outcomes incorrectly, especially when order matters.
• Assuming small experiments must match theoretical probability exactly. They often do not.
• Thinking the Law of Large Numbers means a result is “due.” For example, if a coin lands on Heads several times in a row, that does not mean Tails must happen next.

9. Quick comparison: theoretical vs experimental probability

• Theoretical probability: based on the sample space and math.
• Experimental probability: based on actual trials and data.

When the number of trials is small, these two values may be different.

When the number of trials becomes large, they often become closer.

10. Final summary

A sample space is the complete list of all possible outcomes in a probability experiment. We use it to calculate theoretical probability by comparing favorable outcomes to total outcomes.

Experimental probability comes from actual results. The Law of Large Numbers tells us that as the number of trials increases, experimental probability tends to move closer to theoretical probability.

If you can list outcomes carefully and compare expected results with actual data, you are building a strong foundation in probability.

Mutually Exclusive Events and the Addition Rule

Probability helps us measure how likely something is to happen. In many questions, we want to know the chance that one event or another event happens.

For example, what is the probability of rolling a 2 or a 5 on a number cube? What is the probability of choosing a heart or a king from a deck of cards? To answer questions like these, we use the Addition Rule.

This lesson will teach you how to recognize mutually exclusive events, how to deal with events that overlap, and how to use the correct probability formula.

1. What does “A or B” mean in probability?

In probability, the word or means that either event can happen, or both can happen, unless the events cannot happen at the same time.

So when we write (A or B), we mean the probability that event A happens, event B happens, or both happen.

In symbols, this is written as:

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

2. Mutually exclusive events

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

For example, if you roll one standard number cube:

• Event A: rolling a 2
• Event B: rolling a 5

You cannot roll both a 2 and a 5 on the same roll. So these events are mutually exclusive.

Another example:

• Event A: choosing a red marble
• Event B: choosing a blue marble

If you pick only one marble, it cannot be both red and blue at the same time. These are also mutually exclusive.

When events are mutually exclusive, there is no overlap.

3. The Addition Rule for mutually exclusive events

If two events are mutually exclusive, then:

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

This works because there is no shared outcome being counted twice.

Worked Example 1: Rolling a number cube

Find the probability of rolling a 1 or a 6 on a fair six-sided number cube.

Step 1: Find each probability.

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

Step 2: Check if the events are mutually exclusive.

Yes. One roll cannot be both 1 and 6.

Step 3: Add the probabilities.

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

Answer: $$\frac{1}{3}$$

4. Events that are not mutually exclusive

Sometimes two events can happen at the same time. When this happens, the events are not mutually exclusive.

For example, suppose you choose one card from a standard deck:

• Event A: choosing a heart
• Event B: choosing a king

These events overlap because there is one card that is both a heart and a king: the king of hearts.

If we simply add $$P(A)+P(B)$$, that shared outcome gets counted twice. So we must subtract the overlap once.

5. The general Addition Rule

For any two events A and B:

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

Here:

• $$P(A)$$ is the probability of event A
• $$P(B)$$ is the probability of event B
• $$P(A \text{ and } B)$$ is the probability that both events happen at the same time

This formula is called the Addition Rule.

Important idea: We subtract the intersection because it was counted twice when we added $$P(A)$$ and $$P(B)$$.

6. Visualizing overlap

Imagine two circles that overlap:

• One circle represents event A
• The other circle represents event B

The overlapping part shows outcomes that are in both events. When finding $$P(A \text{ or } B)$$, we want everything in either circle, but we must be careful not to count the overlapping region twice.

Worked Example 2: Choosing a card

A card is chosen at random from a standard 52-card deck. Find the probability of choosing a heart or a king.

Step 1: Find each probability.

• There are 13 hearts, so $$P(\text{heart})=\frac{13}{52}$$
• There are 4 kings, so $$P(\text{king})=\frac{4}{52}$$

Step 2: Find the overlap.

There is 1 card that is both a heart and a king: the king of hearts.

So:

$$P(\text{heart and king})=\frac{1}{52}$$

Step 3: Use the Addition Rule.

$$P(\text{heart or king})=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}$$

$$=\frac{16}{52}=\frac{4}{13}$$

Answer: $$\frac{4}{13}$$

7. How to tell which formula to use

Ask yourself this question:

Can both events happen at the same time?

• If no, the events are mutually exclusive, so use:

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

• If yes, the events are not mutually exclusive, so use:

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

8. Worked Example 3: Spinner with overlap

A spinner has 8 equal sections numbered 1 through 8. Find the probability of spinning an even number or a number greater than 5.

Step 1: List the outcomes in each event.

• Even numbers: $$\{2,4,6,8\}$$
• Numbers greater than 5: $$\{6,7,8\}$$

Step 2: Find the overlap.

The numbers in both sets are:

$$\{6,8\}$$

So the events are not mutually exclusive.

Step 3: Write the probabilities.

• $$P(\text{even})=\frac{4}{8}$$
• $$P(>5)=\frac{3}{8}$$
• $$P(\text{even and } >5)=\frac{2}{8}$$

Step 4: Use the Addition Rule.

$$P(\text{even or } >5)=\frac{4}{8}+\frac{3}{8}-\frac{2}{8}$$

$$=\frac{5}{8}$$

Answer: $$\frac{5}{8}$$

9. Worked Example 4: Using counts from a group

In a class of 30 students:

• 18 play soccer
• 12 play basketball
• 5 play both soccer and basketball

If one student is chosen at random, what is the probability that the student plays soccer or basketball?

Step 1: Use the Addition Rule with counts.

Number who play soccer or basketball:

$$18+12-5=25$$

We subtract 5 because those students were counted in both groups.

Step 2: Write the probability.

$$P(\text{soccer or basketball})=\frac{25}{30}=\frac{5}{6}$$

Answer: $$\frac{5}{6}$$

10. Common mistakes to avoid

• Adding without checking for overlap. If the events can happen together, subtract the intersection.
• Confusing “or” with “and.” “Or” means one event, the other event, or both. “And” means both at the same time.
• Forgetting to simplify. After finding a probability, reduce the fraction if possible.
• Not listing outcomes. If you are unsure whether events overlap, list the outcomes in each event.

11. Quick strategy for solving these problems

1. Identify event A and event B.
2. Decide whether they are mutually exclusive.
3. If they cannot happen together, add the probabilities.
4. If they can happen together, add and then subtract the overlap.
5. Check that your answer makes sense and is between 0 and 1.

12. 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:

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

If the events do overlap, then use the full Addition Rule:

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

The key idea is simple: add the probabilities, but subtract any overlap so you do not count it twice.

Independent Events and the Multiplication Rule

Probability helps us measure how likely something is to happen. In many real-life situations, we want to know the chance that more than one event happens.

For example, what is the probability of flipping a coin and getting heads, and then rolling a die and getting a 4? Questions like this use the idea of independent events and the multiplication rule.

In this lesson, you will learn:

• what independent events are,
• how to tell when events are independent,
• how to use the multiplication rule, and
• how to solve probability problems involving “and”.

1. What is an event?

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

• Getting heads on a coin flip is an event.
• Rolling an even number on a die is an event.
• Drawing a red card from a deck is an event.

Probability is written as a number from 0 to 1, or as a fraction, decimal, or percent.

• Impossible event: probability 0
• Certain event: probability 1
• Example: a fair coin landing heads has probability \(\frac{1}{2}\)

2. What does independent mean?

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

In simpler words, one event does not affect the other.

Examples of independent events:

• Flipping a coin and rolling a die
• Rolling a die, then rolling it again
• Choosing a card, putting it back, shuffling, and choosing again

Examples of events that are not independent:

• Choosing two cards from a deck without replacement
• Taking two marbles from a bag without putting the first one back

In those cases, the first event changes what is available for the second event.

3. The multiplication rule for independent events

When two events \(A\) and \(B\) are independent, the probability that both happen is:

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

This is called the multiplication rule for independent events.

The word “and” is important. It means we want both events to happen in the same situation.

So if two events do not affect each other, you can:

1. Find the probability of the first event.
2. Find the probability of the second event.
3. Multiply the probabilities.

4. Why multiplication makes sense

Suppose you flip a fair coin and roll a fair six-sided die.

The probability of heads is \(\frac{1}{2}\). The probability of rolling a 4 is \(\frac{1}{6}\).

Since flipping the coin does not change the die roll, the events are independent. So:

$$P(\text{heads and 4}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$

This means there is a \(\frac{1}{12}\) chance of getting heads and a 4.

5. Worked Example 1: Coin and die

Question: A fair coin is flipped and a fair die is rolled. What is the probability of getting tails and then a 5?

Step 1: Find each probability.

• Probability of tails: \(\frac{1}{2}\)
• Probability of rolling a 5: \(\frac{1}{6}\)

Step 2: Multiply because the events are independent.

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

Answer: \(\frac{1}{12}\)

6. Worked Example 2: Two spins

Question: A spinner has 4 equal sections: red, blue, green, and yellow. It is spun twice. What is the probability of landing on red both times?

Each spin is independent because the first spin does not change the second spin.

Step 1: Find the probability of red on one spin.

$$P(\text{red}) = \frac{1}{4}$$

Step 2: Multiply for two independent spins.

$$P(\text{red and red}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$$

Answer: \(\frac{1}{16}\)

7. Worked Example 3: More than two independent events

The multiplication rule can also be used for three or more independent events.

Question: A fair coin is flipped 3 times. What is the probability of getting heads all 3 times?

Each flip is independent.

The probability of heads on one flip is \(\frac{1}{2}\).

Multiply three times:

$$P(\text{H, H, H}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

Answer: \(\frac{1}{8}\)

This shows an important idea: when you keep multiplying numbers less than 1, the result gets smaller. That makes sense because getting several specific outcomes in a row is less likely than getting just one.

8. Worked Example 4: Using fractions, decimals, and percents

Question: A student has a \(0.7\) probability of making the first basketball shot and a \(0.7\) probability of making the second shot. If the shots are independent, what is the probability the student makes both shots?

Step 1: Use the multiplication rule.

$$P(\text{make both}) = 0.7 \times 0.7$$ $$= 0.49$$

Answer: The probability is \(0.49\), or \(49\%\).

9. How to tell if events are independent

Ask yourself this question:

Does the first event change the chances for the second event?

• If no, the events are independent.
• If yes, the events are not independent.

Examples:

• Rolling a die twice: independent
• Flipping a coin twice: independent
• Drawing two cards without replacement: not independent
• Drawing a card, replacing it, then drawing again: independent

10. A note about replacement

The phrase with replacement means the item is put back before the next selection. This keeps the probabilities the same, so the events are independent.

The phrase without replacement means the item is not put back. This changes the next probability, so the events are usually not independent.

For example, if you draw a red marble from a bag and do not replace it, there is now one less red marble in the bag. The second draw is affected by the first draw.

11. Common mistakes to avoid

• Adding instead of multiplying: For “and” with independent events, multiply.
• Forgetting to check independence: Do not use the multiplication rule for independent events unless the events really are independent.
• Using the wrong probability: Read carefully. For example, rolling an even number on a die is \(\frac{3}{6}=\frac{1}{2}\), not \(\frac{1}{6}\).
• Not simplifying: If possible, simplify fractions like \(\frac{3}{12}\) to \(\frac{1}{4}\).

12. Quick practice ideas

Try these on your own:

• A coin is flipped twice. What is the probability of getting tails both times?
• A die is rolled twice. What is the probability of getting a 2 and then a 6?
• A spinner with 5 equal sections is spun 3 times. What is the probability of landing on blue all 3 times, if blue is one section?

Answers:

• \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)
• \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
• \(\frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{125}\)

13. Summary

Independent events are events where one outcome does not affect the other. When events are independent, the probability that both happen is found by multiplying their probabilities.

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

Always look for the word “and” and check whether the events are independent. If they are, multiply. This rule works for two events or even more than two events in a row.

Conditional Probability is about finding the chance of an event happening when we already know that another event has happened.

In everyday life, new information can change the chance of something. For example, the chance of picking a red card from a full deck is different from the chance of picking a red card if you already know the card is a heart or a face card. Conditional probability helps us adjust the sample space to match the new information.

We read conditional probability like this: the probability of event A given event B. It is written as \(P(A\mid B)\).

This means: What is the probability that A happens, if we already know B has happened?

Main idea: when we know event \(B\) has happened, we no longer look at all possible outcomes. We only look at the outcomes inside event \(B\).

The formula for conditional probability is:

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

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

This formula only makes sense when \(P(B)\neq 0\), because we cannot divide by 0.

Why do we divide by \(P(B)\)? Because once we know \(B\) has happened, event \(B\) becomes the new “whole” sample space. We are asking what fraction of the outcomes in \(B\) are also in \(A\).

You can also think of it like this:

$$\text{conditional probability} = \frac{\text{part you want inside the new group}}{\text{new group}}$$

So if we know \(B\) happened, then:

• the new group is all outcomes in \(B\)
• the part we want is the outcomes that are in both \(A\) and \(B\)

Important change in denominator: In regular probability, the denominator is usually the total number of possible outcomes. In conditional probability, the denominator changes to the number of outcomes that satisfy the condition.

That is the most important idea in this lesson.

Example 1: Marbles in a bag

A bag has 5 red marbles and 3 blue marbles. One marble is chosen at random.

Let:

• \(A\): the marble is red
• \(B\): the marble is not blue

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

Step 1: Understand event \(B\).

If the marble is not blue, then it must be red. So event \(B\) includes only the 5 red marbles.

Step 2: Restrict the sample space.

Since we know \(B\) happened, we only look at those 5 marbles.

Step 3: Find how many of those also satisfy \(A\).

All 5 are red, so all 5 satisfy \(A\).

Therefore,

$$P(A\mid B)=\frac{5}{5}=1$$

Answer: \(1\), or 100%.

This makes sense because if we already know the marble is not blue, then it must be red.

Example 2: Rolling a die

A fair six-sided die is rolled.

Let:

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

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

Step 1: List event \(B\).

Numbers greater than 3 are \(\{4,5,6\}\).

Step 2: Among those outcomes, find which are even.

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

Step 3: Form the conditional probability.

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

Answer: \(\frac{2}{3}\).

We can also use the formula:

\(A\cap B\) means the outcome is even and greater than 3, so \(A\cap B=\{4,6\}\).

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

Both methods give the same answer.

Example 3: Choosing a card from a deck

A standard deck has 52 cards.

Let:

• \(A\): drawing a queen
• \(B\): drawing a face card

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

Step 1: Count the face cards.

Face cards are jacks, queens, and kings. There are 3 per suit and 4 suits, so there are \(12\) face cards.

Step 2: Count how many of those are queens.

There are 4 queens.

Step 3: Use the new denominator.

Because we know the card is a face card, the sample space is now only the 12 face cards.

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

Answer: \(\frac{1}{3}\).

Notice the difference:

• Without any condition, \(P(\text{queen})=\frac{4}{52}=\frac{1}{13}\)
• Given that the card is a face card, \(P(\text{queen}\mid \text{face card})=\frac{4}{12}=\frac{1}{3}\)

The condition changes the denominator, so the probability changes too.

Example 4: Two draws without replacement

A bag contains 3 green balls and 2 yellow balls. Two balls are drawn, one after the other, without replacement.

Find the probability that the second ball is green, given that the first ball was yellow.

Let:

• \(A\): second ball is green
• \(B\): first ball is yellow

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

Step 1: Use the condition.

If the first ball was yellow, then one yellow ball has been removed.

Now the bag has:

• 3 green balls
• 1 yellow ball

So there are 4 balls left in total.

Step 2: Find the chance that the second ball is green from this new situation.

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

Answer: \(\frac{3}{4}\).

This is a good example of a dependent situation. The first draw changes what happens on the second draw because the ball is not replaced.

How to solve conditional probability problems

1. Identify the event you want, \(A\).
2. Identify the condition, \(B\).
3. Change the sample space so it includes only outcomes in \(B\).
4. Count how many outcomes in \(B\) also satisfy \(A\).
5. Write the probability as: $$P(A\mid B)=\frac{\text{outcomes in both }A\text{ and }B}{\text{outcomes in }B}$$

A common mistake is forgetting to change the denominator.

For example, if you are asked for the probability of drawing a queen given that the card is a face card, the denominator is 12, not 52. Once the condition is given, the “total possible” outcomes become only those that fit the condition.

Another useful relationship

We can rearrange the conditional probability formula to get:

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

This says:

• probability that \(B\) happens,
• times the probability that \(A\) happens once \(B\) has happened,
• equals the probability that both happen.

This can help when working backward from a conditional probability question.

Quick check

• If you know an event has happened, ask: What is my new sample space?
• The denominator should count only outcomes in the condition.
• The numerator should count outcomes that satisfy both the condition and the event you want.

Summary

Conditional probability tells us the chance of an event when another event is already known to have happened. The key idea is that the sample space changes, so the denominator changes too.

The formula is:

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

When solving these problems, always focus on the condition first. Once you know the new group of possible outcomes, it becomes much easier to find the correct probability.

Combinations and Permutations help us count how many different outcomes are possible in a situation.

They are very useful in probability, because before we can find the chance of something happening, we often need to know how many total arrangements or selections are possible.

The big idea is this:

• Permutation: use this when order matters.
• Combination: use this when order does not matter.

For example, choosing a president, vice president, and secretary from a group is a permutation because each position is different.

Choosing 3 students to be on a team is a combination because it does not matter which student is listed first.

To understand both ideas, we first need to know about factorials.

Factorial means multiplying a whole number by all the whole numbers below it down to 1.

We write factorial with an exclamation mark.

For example:

• \(4! = 4 \times 3 \times 2 \times 1 = 24\)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
• \(1! = 1\)

A very important fact is:

$$0! = 1$$

This may seem strange at first, but it makes the formulas work correctly.

Permutations count arrangements when order matters.

If we want to arrange all objects in a set, we use factorial.

For example, the number of ways to arrange 3 books on a shelf is:

$$3! = 6$$

The 6 arrangements are:

• ABC
• ACB
• BAC
• BCA
• CAB
• CBA

If we are only choosing and arranging part of a group, we use the permutation formula:

$$P(n,r) = \frac{n!}{(n-r)!}$$

Here:

• \(n\) = total number of items
• \(r\) = number of items chosen

This formula counts how many ways we can choose and arrange \(r\) items from \(n\) items.

Combinations count selections when order does not matter.

If we choose 2 fruits, apple then banana is the same choice as banana then apple. Since the order does not create a new group, we use a combination.

The combination formula is:

$$C(n,r) = \frac{n!}{r!(n-r)!}$$

You may also see it written as \(\binom{n}{r}\).

This formula counts how many ways we can choose \(r\) items from \(n\) items when order does not matter.

Here is a helpful way to decide which one to use:

• If changing the order creates a different outcome, use permutation.
• If changing the order does not create a different outcome, use combination.

Worked Example 1: Simple permutation

How many ways can 4 different students stand in a line?

Since the students are being arranged in order, order matters.

So we use factorial:

$$4! = 4 \times 3 \times 2 \times 1 = 24$$

Answer: There are \(24\) ways.

Worked Example 2: Permutation of part of a group

From 6 runners, how many ways can 1st place, 2nd place, and 3rd place be awarded?

The order matters because 1st, 2nd, and 3rd are different positions.

So we use a permutation:

$$P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}$$

Now simplify:

$$\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 6 \times 5 \times 4 = 120$$

Answer: There are \(120\) ways to award the 3 places.

Worked Example 3: Simple combination

A class has 8 students. How many different groups of 3 students can be chosen?

This is a selection, not an arrangement. A group with students A, B, and C is the same group as C, A, and B.

So order does not matter, and we use a combination:

$$C(8,3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}$$

Simplify by canceling:

$$C(8,3) = \frac{8 \times 7 \times 6 \times 5!}{(3 \times 2 \times 1)5!}$$

$$= \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56$$

Answer: There are \(56\) different groups.

Worked Example 4: Comparing permutation and combination

From 5 books, how many ways can you choose 2 books if:

1. the books are arranged on a shelf,
2. the books are just chosen to take home?

Part 1: Arranged on a shelf

Now order matters, because choosing book A then B is different from B then A on the shelf.

Use permutation:

$$P(5,2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$$

Part 2: Chosen to take home

Now order does not matter. The same 2 books are the same choice no matter how we list them.

Use combination:

$$C(5,2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!}$$

$$= \frac{5 \times 4 \times 3!}{(2 \times 1)3!} = \frac{20}{2} = 10$$

Answer:

• Arranged on a shelf: \(20\) ways
• Chosen only: \(10\) ways

Notice that the permutation answer is larger because it counts different orders separately.

Why does the combination formula divide by \(r!\)?

When we use permutations, we count every possible order of the chosen items.

But in combinations, different orders of the same group should count as only one group.

For example, choosing 3 students A, B, and C gives these 6 orders:

• ABC
• ACB
• BAC
• BCA
• CAB
• CBA

Since \(3! = 6\), we divide by \(3!\) to remove the repeated counting of the same group.

That is why:

$$C(n,r) = \frac{P(n,r)}{r!} = \frac{n!}{r!(n-r)!}$$

Common mistakes to avoid

• Using a permutation when the question is only about choosing a group.
• Using a combination when the question gives positions like 1st, 2nd, 3rd.
• Forgetting that factorial means multiplying all whole numbers down to 1.
• Not simplifying carefully when working with factorials.

Quick check questions

• How many ways can 3 letters be arranged? \(3! = 6\)
• How many ways can 2 winners be chosen from 7 students if there is a 1st and 2nd prize? Use \(P(7,2)\).
• How many ways can 2 students be chosen from 7 students for a team? Use \(C(7,2)\).

Summary

Permutations and combinations are both ways to count possible outcomes.

Use a permutation when order matters:

$$P(n,r) = \frac{n!}{(n-r)!}$$

Use a combination when order does not matter:

$$C(n,r) = \frac{n!}{r!(n-r)!}$$

Always ask yourself one question first: Does the order change the outcome? If yes, use permutations. If no, use combinations.

Expected Value is a way to find the long-term average outcome of a situation involving chance.

It does not tell you exactly what will happen in one try. Instead, it tells you what you would expect on average if the same experiment or game happened many times.

For example, if a game gives different prizes with different probabilities, expected value helps answer questions like:

• Is this game fair?
• How much money would I win or lose on average?
• Which choice is better in the long run?

This idea is useful in games, business, insurance, and finance because it helps compare risk and reward.

1. The basic idea

Suppose a random event can have several outcomes. Each outcome has:

• a value, such as money won or lost
• a probability, which is how likely it is to happen

To find expected value, multiply each outcome by its probability, then add the results.

The formula is:

$$EV = \sum (\text{outcome} \times \text{probability})$$

Written more fully:

$$EV = x_1p_1 + x_2p_2 + x_3p_3 + \cdots$$

where:

• \(x_1, x_2, x_3, \dots\) are the possible outcomes

• \(p_1, p_2, p_3, \dots\) are their probabilities

The probabilities should add up to 1:

$$p_1 + p_2 + p_3 + \cdots = 1$$

2. What expected value means

If the expected value of a game is \(2\), that means the average result is 2 units per play over many plays.

If the expected value is negative, such as \(-1.50\), that means you would lose an average of $1.50 per play in the long run.

Expected value can even be a number that is not one of the possible outcomes.

For example, if you flip a coin and win $0 or $1, the expected value is \(0.5\) dollars. You never actually win 50 cents in one flip, but 50 cents is the average amount per flip over many flips.

3. Steps for finding expected value

1. List all possible outcomes.

2. Write the probability of each outcome.

3. Multiply each outcome by its probability.

4. Add the products.

4. Worked Example 1: Rolling a fair die

A fair six-sided die has outcomes 1, 2, 3, 4, 5, and 6. Each outcome has probability \(\frac{1}{6}\).

Find the expected value.

Step 1: Write the formula

$$EV = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right)$$

Step 2: Combine the terms

$$EV = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$$

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

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

5. Worked Example 2: A simple prize game

You spin a wheel.

• You win $10 with probability \(\frac{1}{4}\)

• You win $4 with probability \(\frac{1}{2}\)

• You win $0 with probability \(\frac{1}{4}\)

Find the expected value.

Step 1: Multiply each prize by its probability

$$10\left(\frac{1}{4}\right) = 2.5$$ $$4\left(\frac{1}{2}\right) = 2$$ $$0\left(\frac{1}{4}\right) = 0$$

Step 2: Add the results

$$EV = 2.5 + 2 + 0 = 4.5$$

Answer: The expected value is \(\$4.50\).

This means the average amount won per spin is $4.50 in the long run.

6. Expected value with profit or loss

Often, expected value is used to decide whether a game is fair.

A game is fair if the expected gain is 0. In other words, neither side has an advantage in the long run.

If you must pay to play, then you should include that cost when finding your expected profit.

Expected profit can be found by:

$$\text{Expected profit} = \text{Expected winnings} - \text{Cost to play}$$

7. Worked Example 3: Is the game fair?

A game works like this:

• If you draw a red card, you win $8.

• If you draw a black card, you win $0.

• You pay $5 to play.

Assume the card is drawn from a standard deck, so red and black are each equally likely. That means:

$$P(\text{red}) = \frac{1}{2}, \qquad P(\text{black}) = \frac{1}{2}$$

Step 1: Find expected winnings

$$EV = 8\left(\frac{1}{2}\right) + 0\left(\frac{1}{2}\right) = 4$$

So the expected winnings are \(\$4\).

Step 2: Subtract the cost to play

$$\text{Expected profit} = 4 - 5 = -1$$

Answer: The expected profit is \(-$1\).

This means that on average, a player loses $1 each time. So the game is not fair for the player.

8. Worked Example 4: Using a probability table

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

Outcome \(x\)	Probability \(P(X=x)\)
0	0.2
2	0.5
7	0.3

Find \(E(X)\), the expected value of \(X\).

Step 1: Multiply each outcome by its probability

$$0(0.2) = 0$$ $$2(0.5) = 1$$ $$7(0.3) = 2.1$$

Step 2: Add the results

$$E(X) = 0 + 1 + 2.1 = 3.1$$

Answer: The expected value is \(3.1\).

9. Common mistakes to avoid

• Forgetting to multiply by probability
  Do not just average the outcomes unless they are all equally likely.

• Using probabilities that do not add to 1
  Check that all probabilities together make a complete set of outcomes.

• Forgetting the cost of playing
  If a game charges money to play, include that when finding expected profit.

• Thinking expected value must be a possible outcome
  It can be a number like 3.5 or 2.1, even if that exact result never happens in one trial.

10. When outcomes are equally likely

If all outcomes are equally likely, expected value is just the average of the outcomes.

For example, for a fair die:

$$EV = \frac{1+2+3+4+5+6}{6}$$

This works because each outcome has the same probability.

But if probabilities are different, you must use a weighted average. That means outcomes with higher probability count more.

11. Why expected value matters

Expected value helps people make decisions.

• In games, it shows whether a game favors the player or the organizer.

• In finance, it helps compare choices with different risks and rewards.

• In everyday life, it helps you think logically about chances instead of guessing.

12. Quick check

Try these on your own:

1. A bag has 3 red marbles and 1 blue marble. You win $6 for red and $0 for blue. What is the expected value?

2. A game pays $20 with probability \(0.1\) and $2 with probability \(0.9\). What is the expected value?

3. A raffle ticket costs $3. You have a \(\frac{1}{5}\) chance to win $10 and a \(\frac{4}{5}\) chance to win nothing. What is your expected profit?

Answers:

1. $$EV = 6\left(\frac{3}{4}\right) + 0\left(\frac{1}{4}\right) = 4.5$$

2. $$EV = 20(0.1) + 2(0.9) = 2 + 1.8 = 3.8$$

3. Expected winnings: $$10\left(\frac{1}{5}\right)+0\left(\frac{4}{5}\right)=2$$ Expected profit: $$2-3=-1$$

Summary

Expected value is the long-term average result of a random situation. To find it, multiply each outcome by its probability and add the results.

It is especially useful for judging games, prizes, and financial choices. A positive expected value means an average gain, a negative expected value means an average loss, and zero means the situation is fair.

Probability Distributions Basics is about organizing all the possible outcomes of a chance experiment and showing the probability of each outcome.

Instead of finding just one probability, a probability distribution lists every possible value of a random result and how likely each value is.

This is useful because it helps us see the full picture of what can happen in an experiment, game, or real-life situation.

Example idea: If you flip two coins, you might want to know the probability of getting 0 heads, 1 head, or 2 heads. A probability distribution organizes those results neatly.

1. What is a probability distribution?

A probability distribution for a discrete random variable is a table, list, or graph that shows:

• all possible values of the variable, and
• the probability of each value.

A random variable is a number that represents the outcome of a chance experiment.

For 9th Grade, we will focus on discrete random variables. That means the possible values are separate, countable numbers like 0, 1, 2, 3, and so on.

Example: Let \(X\) be the number of heads when flipping two coins. Then the possible values of \(X\) are:

$$X = 0, 1, 2$$

Each of these values has a probability, so together they make a probability distribution.

2. Important rules for a probability distribution

A correct probability distribution must follow two main rules:

1. Each probability must be between 0 and 1.
2. All the probabilities must add up to 1.

In symbols, if the possible values are \(x_1, x_2, x_3, \dots\), then:

$$0 \le P(X=x) \le 1$$

and

$$P(x_1)+P(x_2)+P(x_3)+\cdots = 1$$

This makes sense because probabilities cannot be negative, cannot be more than 1, and all possible outcomes together must account for 100% of what can happen.

3. How to make a probability distribution

To create a probability distribution, follow these steps:

1. Decide what the random variable represents.
2. List all possible values of the variable.
3. Find the probability of each value.
4. Check that all probabilities add to 1.

It often helps to first list the actual outcomes of the experiment, then group them by the value of the random variable.

Worked Example 1: Rolling one fair die

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

Step 1: List possible values.

$$X = 1, 2, 3, 4, 5, 6$$

Step 2: Find each probability.

Because the die is fair, each number is equally likely.

$$P(X=1)=P(X=2)=P(X=3)=P(X=4)=P(X=5)=P(X=6)=\frac{1}{6}$$

Probability distribution:

• \(P(X=1)=\frac{1}{6}\)
• \(P(X=2)=\frac{1}{6}\)
• \(P(X=3)=\frac{1}{6}\)
• \(P(X=4)=\frac{1}{6}\)
• \(P(X=5)=\frac{1}{6}\)
• \(P(X=6)=\frac{1}{6}\)

Check:

$$\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=1$$

So this is a valid probability distribution.

4. Probability distributions can be shown in a table

A table is one of the clearest ways to show a probability distribution.

For the die example:

Value of \(X\): 1, 2, 3, 4, 5, 6

Probability: \(\frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}\)

You can also make a bar graph where the horizontal axis shows the values of \(X\) and the vertical axis shows the probabilities.

5. Worked Example 2: Tossing two coins

Let \(X\) be the number of heads in two coin tosses.

Step 1: List all equally likely outcomes.

$$HH, HT, TH, TT$$

There are 4 outcomes.

Step 2: Find the value of \(X\) for each outcome.

• \(HH \rightarrow X=2\)
• \(HT \rightarrow X=1\)
• \(TH \rightarrow X=1\)
• \(TT \rightarrow X=0\)

Step 3: Find probabilities.

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

Probability distribution:

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

Check:

$$\frac{1}{4}+\frac{1}{2}+\frac{1}{4}=1$$

This distribution shows that getting 1 head is the most likely result.

6. Not all distributions are uniform

A uniform distribution means all values have the same probability, like rolling one fair die.

But many probability distributions are not uniform. In the two-coin example, the probabilities are not all equal.

This happens because even if the basic outcomes are equally likely, the values of the random variable may not happen equally often.

7. Worked Example 3: Sum of two dice

Let \(X\) be the sum when rolling two fair dice.

Step 1: Find the possible values.

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

$$X = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$

Step 2: Count how many ways each sum can happen.

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

Since two dice have:

$$6 \times 6 = 36$$

total equally likely outcomes, each probability is:

$$P(X=\text{sum})=\frac{\text{number of ways}}{36}$$

Some values in the distribution:

• \(P(X=2)=\frac{1}{36}\)
• \(P(X=7)=\frac{6}{36}=\frac{1}{6}\)
• \(P(X=12)=\frac{1}{36}\)

This shows why 7 is the most likely sum.

8. Reading a probability distribution

Once a distribution is given, you should be able to answer questions from it.

For example, in the two-coin distribution:

• Probability of exactly 2 heads: \(P(X=2)=\frac{1}{4}\)
• Probability of at least 1 head: \(P(X\ge 1)=P(X=1)+P(X=2)\)

So:

$$P(X\ge 1)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$$

The phrase at least means the value is that number or more.

The phrase at most means the value is that number or less.

9. Worked Example 4: Using a distribution to answer questions

A bag contains 3 red marbles and 1 blue marble. One marble is picked at random. Let \(X\) represent the number of blue marbles picked.

Since only one marble is chosen, the possible values are:

$$X=0 \text{ or } 1$$

Step 1: Find probabilities.

• \(P(X=1)=\frac{1}{4}\) because there is 1 blue marble out of 4 total.
• \(P(X=0)=\frac{3}{4}\) because there are 3 non-blue outcomes out of 4.

Probability distribution:

• \(P(X=0)=\frac{3}{4}\)
• \(P(X=1)=\frac{1}{4}\)

Question: What is the probability of picking at most 1 blue marble?

Because the only possible values are 0 and 1, “at most 1” includes both of them.

$$P(X\le 1)=P(X=0)+P(X=1)=\frac{3}{4}+\frac{1}{4}=1$$

That makes sense, because every outcome is either 0 blue marbles or 1 blue marble.

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

If you are given a table of values and probabilities, ask these questions:

• Are all probabilities between 0 and 1?
• Do all probabilities add up to 1?
• Does the variable use clear possible values?

Example check:

Suppose a table says:

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

Check the sum:

$$0.2+0.5+0.3=1.0$$

All probabilities are between 0 and 1, so this is a valid probability distribution.

But if the probabilities added to 1.2 or included a negative number, the table would not be valid.

11. Common mistakes to avoid

• Forgetting possible values: Make sure you list every value the random variable can take.
• Mixing outcomes and variable values: The outcome might be \(HT\), but the random variable value might be \(1\).
• Not checking the total: The probabilities must add to 1.
• Assuming all values are equally likely: They are not always equal, even if the original outcomes are equally likely.

12. Why probability distributions matter

Probability distributions help us describe chance in an organized way.

They are used in games, science, weather prediction, sports, and decision-making. When you know the full distribution, you understand not just one outcome, but the whole set of possibilities.

Brief Summary

A probability distribution shows all possible values of a discrete random variable and the probability of each one.

To make one, list the possible values, find each probability, and check that all probabilities are between 0 and 1 and add up to 1.

Probability distributions help you organize outcomes clearly and answer questions like “exactly,” “at least,” and “at most” for a chance experiment.