Probability

Classical Definition of Probability and Sample Spaces

Probability is a way of measuring how likely something is to happen. In everyday life, we use probability when we talk about the chance of rain, winning a game, or drawing a certain card from a deck.

In this lesson, you will learn how to use the classical definition of probability. This method works when all outcomes are equally likely. You will also learn how to describe all possible outcomes using a sample space.

These ideas are important because they help us answer questions like:

• What is the probability of rolling an even number on a die?
• What is the probability of drawing a red card from a deck?
• What is the probability of getting two heads when flipping two coins?

1. What is a sample space?

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

An experiment in probability is any action with a result that can be observed, such as tossing a coin, rolling a die, or choosing a card.

We often use the letter \(S\) to represent the sample space.

Here are some simple sample spaces:

• For flipping one coin: \(S = \{H, T\}\)
• For rolling one die: \(S = \{1,2,3,4,5,6\}\)
• For choosing a day of the weekend: \(S = \{\text{Saturday}, \text{Sunday}\}\)

To solve probability questions correctly, you must list the sample space carefully and make sure no possible outcome is missing.

2. Equally likely outcomes

The classical definition of probability only works when each outcome in the sample space is equally likely.

For example:

• In a fair coin toss, heads and tails are equally likely.
• In a fair die roll, each number from 1 to 6 is equally likely.
• In a well-shuffled deck of cards, each card is equally likely to be chosen.

If outcomes are not equally likely, then we cannot use the classical definition directly.

3. Classical definition of probability

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

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

This can also be written as:

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

Here:

• \(E\) means the event
• \(n(E)\) is the number of favorable outcomes
• \(n(S)\) is the total number of outcomes in the sample space

A probability is always between 0 and 1:

• \(0\) means the event is impossible
• \(1\) means the event is certain
• A value between 0 and 1 shows how likely the event is

Probabilities can also be written as fractions, decimals, or percentages.

For example:

• \(\frac{1}{2} = 0.5 = 50\%\)
• \(\frac{1}{6} \approx 0.167 = 16.7\%\)

4. How to find probability step by step

1. Identify the experiment.
2. List the sample space \(S\).
3. Identify the event \(E\).
4. Count the favorable outcomes.
5. Count the total outcomes.
6. Use the formula \(P(E) = \frac{n(E)}{n(S)}\).
7. Simplify the fraction if possible.

Worked Example 1: Rolling one die

Question: What is the probability of rolling a number greater than 4 on a fair six-sided die?

Step 1: Write the sample space.

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

Step 2: Find the favorable outcomes.

Numbers greater than 4 are \(5\) and \(6\).

So the event is \(E = \{5,6\}\).

Step 3: Count outcomes.

• Number of favorable outcomes: \(n(E) = 2\)
• Total number of outcomes: \(n(S) = 6\)

Step 4: Apply the formula.

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

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

Worked Example 2: Choosing a card

Question: A card is chosen from a standard deck of 52 cards. What is the probability of drawing a heart?

Step 1: Understand the sample space.

A standard deck has 52 cards in total.

Step 2: Count favorable outcomes.

There are 13 hearts in the deck.

Step 3: Use the formula.

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

Answer: The probability of drawing a heart is \(\frac{1}{4}\).

5. Listing sample spaces for more than one action

Sometimes an experiment has more than one step, such as flipping two coins or rolling two dice. In these cases, the sample space must include every possible combination.

Two coin tosses

If you toss two coins, the sample space is:

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

Notice that \(HT\) and \(TH\) are different outcomes because the order matters. The first letter shows the first coin, and the second letter shows the second coin.

Two dice rolls

If two dice are rolled, each outcome is written as an ordered pair:

$$ (1,1), (1,2), (1,3), \dots, (6,6) $$

There are:

$$6 \times 6 = 36$$

possible outcomes in the sample space.

Worked Example 3: Flipping two coins

Question: What is the probability of getting exactly one head when two fair coins are tossed?

Step 1: List the sample space.

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

Step 2: Identify the favorable outcomes.

Exactly one head happens in:

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

Step 3: Count outcomes.

• \(n(E) = 2\)
• \(n(S) = 4\)

Step 4: Apply the formula.

$$P(E) = \frac{2}{4} = \frac{1}{2}$$

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

6. Events and favorable outcomes

An event is a set of outcomes from the sample space.

For example, when rolling one die:

• The event “roll an even number” is \(\{2,4,6\}\)
• The event “roll a number less than 3” is \(\{1,2\}\)
• The event “roll a 7” is \(\{\}\), which is impossible

The favorable outcomes are simply the outcomes in the event you want.

7. Probability of compound events using the sample space

A compound event involves more than one action or more than one condition.

Examples:

• Rolling two dice and finding the sum
• Flipping two coins and checking how many heads appear
• Choosing a card and asking whether it is red and a face card

When solving these, the safest method is to list the sample space carefully, then count the favorable outcomes.

Worked Example 4: Rolling two dice

Question: Two fair dice are rolled. What is the probability that the sum is 8?

Step 1: Total outcomes.

Each die has 6 possible outcomes, so the sample space has:

$$6 \times 6 = 36$$

total outcomes.

Step 2: Find the favorable outcomes.

The pairs that give a sum of 8 are:

$$ (2,6), (3,5), (4,4), (5,3), (6,2) $$

There are 5 favorable outcomes.

Step 3: Apply the formula.

$$P(\text{sum of }8) = \frac{5}{36}$$

Answer: The probability is \(\frac{5}{36}\).

8. Important ideas to remember

• The classical definition of probability only works when outcomes are equally likely.
• A sample space must include all possible outcomes.
• Favorable outcomes are the outcomes that match the event.
• Use:

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

Make sure you count carefully. Many mistakes happen because students forget outcomes or count them incorrectly.

9. Common mistakes

• Forgetting outcomes: For two coin tosses, writing only \(\{HH, HT, TT\}\) and forgetting \(TH\).
• Not treating order correctly: In two dice rolls, \((2,5)\) and \((5,2)\) are different outcomes.
• Using the formula when outcomes are not equally likely: The classical rule needs fairness or equal chance.
• Counting favorable outcomes incorrectly: Read the event carefully before counting.

10. Quick check

Try these on your own:

• What is the probability of drawing a black card from a standard deck?
• What is the probability of getting tails when flipping one fair coin?
• Two coins are tossed. What is the probability of getting at least one head?

Answers:

• Black cards: \(\frac{26}{52} = \frac{1}{2}\)
• Tails: \(\frac{1}{2}\)
• At least one head: favorable outcomes are \(HH, HT, TH\), so \(\frac{3}{4}\)

Summary

A sample space is the full set of possible outcomes in a probability experiment. An event is a group of outcomes you are interested in.

When all outcomes are equally likely, the classical definition of probability is:

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

To solve probability questions well, always list the sample space carefully, identify the favorable outcomes, and count accurately.

Complementary, Certain, and Impossible Events are important ideas in probability. They help us understand what can happen, what cannot happen, and how to work with the opposite of an event.

In probability, every event has a value between 0 and 1.

$$0 \leq P(\text{event}) \leq 1$$

A probability of 0 means the event cannot happen. A probability of 1 means the event must happen. Probabilities in between 0 and 1 describe events that may happen.

These ideas are useful because sometimes it is easier to find the probability that something does not happen, and then use that to find the answer we want.

1. The probability scale

• Impossible event: an event that has probability 0.
• Certain event: an event that has probability 1.
• Possible event: an event with probability between 0 and 1.

For example, when rolling a standard 6-sided die:

• Getting a 7 is impossible, so \(P(7)=0\).
• Getting a number less than 7 is certain, so \(P(\text{number} < 7)=1\).
• Getting an even number is possible, and its probability is between 0 and 1.

2. What is a complementary event?

The complement of an event is the event not happening. If event \(A\) happens, then its complement, written as \(A'\) or \(\text{not }A\), does not happen.

For example:

• If \(A\) is “rolling an even number,” then \(A'\) is “not rolling an even number,” which means rolling an odd number.
• If \(A\) is “choosing a red card,” then \(A'\) is “not choosing a red card.”

An event and its complement cover all possible outcomes. One of them must happen, and they cannot both happen at the same time.

Because of this, their probabilities add to 1:

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

This is called the complement rule.

We can rearrange it in two helpful ways:

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

3. Why the complement rule is helpful

Sometimes finding the probability of an event directly is difficult. But finding the probability of its complement is easier.

For example, suppose you want the probability of getting at least one head when tossing a coin several times. Counting all the ways to get at least one head can be messy. But the complement is simple: no heads, which means all tails.

Then you can use:

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

This saves time and reduces mistakes.

4. Certain and impossible events in sample spaces

A sample space is the set of all possible outcomes.

For one roll of a die, the sample space is:

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

A certain event includes every outcome in the sample space. An impossible event includes none of the outcomes in the sample space.

Example with a die:

• Certain event: “rolling a number from 1 to 6”
• Impossible event: “rolling a 0”

This connects directly to probability:

• If an event has no outcomes, its probability is 0.
• If an event has all outcomes, its probability is 1.

5. Worked Examples

Example 1: Identifying certain and impossible events

A bag contains blue, green, and yellow counters only. You pick one counter at random.

Find the probability of each event:

1. Picking a blue, green, or yellow counter
2. Picking a red counter

Solution

The bag only has blue, green, and yellow counters.

• Picking blue, green, or yellow must happen, so this is a certain event.

$$P(\text{blue, green, or yellow})=1$$

• Picking red cannot happen, so this is an impossible event.

$$P(\text{red})=0$$

Example 2: Finding a complement

A fair die is rolled once. Let \(A\) be the event “roll a number greater than 4.”

Find:

1. \(P(A)\)
2. \(P(A')\)

Solution

The numbers greater than 4 are 5 and 6, so there are 2 favorable outcomes out of 6.

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

The complement \(A'\) means “not greater than 4,” so the outcomes are 1, 2, 3, and 4.

$$P(A')=\frac{4}{6}=\frac{2}{3}$$

We can also check using the complement rule:

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

Example 3: Using the complement rule to solve a problem

A card is chosen at random from a standard deck of 52 cards. What is the probability that the card is not a heart?

Solution

There are 13 hearts in a deck of 52 cards.

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

The event “not a heart” is the complement of “heart.”

$$P(\text{not a heart})=1-P(\text{heart})$$ $$P(\text{not a heart})=1-\frac{1}{4}=\frac{3}{4}$$

So the probability is:

$$\frac{3}{4}$$

Example 4: A more challenging complement problem

A coin is tossed 3 times. What is the probability of getting at least one head?

Solution

The phrase “at least one head” means 1 head, 2 heads, or 3 heads. Instead of listing all of those outcomes, use the complement.

The complement of “at least one head” is “no heads,” which means all tails.

For one fair coin toss:

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

For 3 tails in a row:

$$P(\text{no heads})=P(TTT)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$$

Now use the complement rule:

$$P(\text{at least one head})=1-P(\text{no heads})$$ $$P(\text{at least one head})=1-\frac{1}{8}=\frac{7}{8}$$

So the probability of getting at least one head is:

$$\frac{7}{8}$$

6. Common mistakes to avoid

• Forgetting that probabilities must be between 0 and 1. A probability cannot be negative or greater than 1.
• Mixing up an event and its complement. If \(A\) is “getting a multiple of 3,” then \(A'\) is “not getting a multiple of 3,” not just one specific outcome.
• Not recognizing certain or impossible events. Always compare the event to the full sample space.
• Forgetting to subtract from 1. When using complements, the final step is often \(1 - P(\text{complement})\).

7. Key ideas to remember

• Every probability is between 0 and 1.
• An impossible event has probability 0.
• A certain event has probability 1.
• The complement of an event means the event does not happen.
• An event and its complement add to 1:

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

This rule is especially useful when the complement is easier to calculate than the original event.

Brief Summary

In probability, values always stay between 0 and 1. An impossible event has probability 0, and a certain event has probability 1. The complement of an event is the event not happening, and the probabilities of an event and its complement always add to 1. Using complements is a smart way to solve many probability problems more quickly.

Probability in Coin Tossing Experiments

Probability helps us measure how likely something is to happen. In coin tossing experiments, probability is based on the idea that each possible outcome is equally likely when the coin is fair.

In this lesson, you will learn how to list outcomes, build sample spaces, use tree diagrams, and find the probability of events when two or three coins are tossed.

1. What is probability?

Probability is a number between 0 and 1 that tells how likely an event is.

• Impossible event: probability = 0

• Certain event: probability = 1

• Events in between: probability is between 0 and 1

For equally likely outcomes, we use the formula

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

For a single fair coin toss, the sample space is:

$$\{H, T\}$$

Here, H means heads and T means tails.

Since there are 2 equally likely outcomes,

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

2. Sample space for tossing two coins

When two coins are tossed, we look at the outcome of the first coin and the second coin together.

The sample space is:

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

Each letter shows the result of one coin. For example:

• \(HH\): both coins show heads

• \(HT\): first coin is heads, second coin is tails

• \(TH\): first coin is tails, second coin is heads

• \(TT\): both coins show tails

There are 4 equally likely outcomes, so each outcome has probability

$$\frac{1}{4}$$

3. Using a tree diagram for two coins

A tree diagram helps organize all possible outcomes step by step.

Start with the first coin. It can be H or T. From each of those, the second coin can also be H or T.

This gives:

• First coin H, then second coin H gives \(HH\)

• First coin H, then second coin T gives \(HT\)

• First coin T, then second coin H gives \(TH\)

• First coin T, then second coin T gives \(TT\)

A tree diagram is useful because it makes sure you do not miss any outcomes.

4. Finding probabilities with two coins

Once the sample space is listed, count the outcomes that match the event.

For example, the event “exactly one head” means one head and one tail. The outcomes are:

$$\{HT, TH\}$$

So the probability is

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

Worked Example 1

Question: Two fair coins are tossed. What is the probability of getting both heads?

Step 1: Write the sample space.

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

Step 2: Find the favorable outcomes.

“Both heads” means only \(HH\).

Step 3: Use the probability formula.

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

Answer: The probability is \(\frac{1}{4}\).

Worked Example 2

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

Step 1: Write the sample space.

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

Step 2: Find the outcomes with at least one tail.

These are:

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

Step 3: Count and calculate.

There are 3 favorable outcomes out of 4 total outcomes.

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

Answer: The probability is \(\frac{3}{4}\).

5. Sample space for tossing three coins

Now let us extend the idea to three coins.

Each coin has 2 possible outcomes, so the total number of outcomes is

$$2 \times 2 \times 2 = 8$$

The sample space is:

$$\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$

Each outcome is equally likely, so each has probability

$$\frac{1}{8}$$

6. Using a tree diagram for three coins

For three coins, the tree has three stages:

1. First coin: H or T

2. Second coin: H or T from each branch

3. Third coin: H or T from each new branch

Following all branches gives 8 final outcomes. A tree diagram is especially helpful for three coins because the sample space is larger and easier to mix up.

7. Events involving number of heads

In coin tossing, many questions ask about the number of heads.

For three coins:

• 0 heads: \(TTT\)

• 1 head: \(HTT, THT, TTH\)

• 2 heads: \(HHT, HTH, THH\)

• 3 heads: \(HHH\)

This grouping helps you answer questions quickly.

Worked Example 3

Question: Three fair coins are tossed. What is the probability of getting exactly two heads?

Step 1: Write the sample space.

$$\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$

Step 2: Find the outcomes with exactly two heads.

These are:

$$\{HHT, HTH, THH\}$$

Step 3: Calculate the probability.

$$P(\text{exactly two heads}) = \frac{3}{8}$$

Answer: The probability is \(\frac{3}{8}\).

Worked Example 4

Question: Three fair coins are tossed. What is the probability of getting at least two heads?

Step 1: Understand “at least two heads.”

This means either exactly two heads or exactly three heads.

Step 2: List the favorable outcomes.

Exactly two heads: \(HHT, HTH, THH\)

Exactly three heads: \(HHH\)

So the favorable outcomes are:

$$\{HHT, HTH, THH, HHH\}$$

Step 3: Calculate the probability.

$$P(\text{at least two heads}) = \frac{4}{8} = \frac{1}{2}$$

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

8. Important words in probability questions

Some words appear often in coin toss questions. Make sure you know what they mean.

• Exactly means the precise number only.

• At least means that number or more.

• At most means that number or less.

• Both means the two coins show the same stated result.

• All means every coin shows the stated result.

For example, with three coins:

• Exactly one head: only 1 head

• At least one head: 1, 2, or 3 heads

• At most one head: 0 or 1 head

9. Common mistakes to avoid

• Forgetting outcomes: Always list the full sample space or use a tree diagram.

• Mixing up HT and TH: These are different outcomes because the order matters.

• Misreading words: “Exactly two” is not the same as “at least two.”

• Using the wrong total: Two coins have 4 outcomes, and three coins have 8 outcomes.

10. Quick method to count total outcomes

Each coin has 2 possible outcomes. If you toss:

• 1 coin: \(2\) outcomes

• 2 coins: \(2^2 = 4\) outcomes

• 3 coins: \(2^3 = 8\) outcomes

So for coin tossing, the total number of outcomes is found by multiplying 2 for each coin tossed.

11. Summary

In coin tossing experiments, probability is found by comparing favorable outcomes to total possible outcomes. For two coins, there are 4 equally likely outcomes, and for three coins, there are 8 equally likely outcomes.

Tree diagrams and sample spaces help you organize outcomes clearly. Once you list all possible results, you can answer questions about events such as exactly one head, both tails, or at least two heads.

The key is to work carefully, count accurately, and pay attention to words like exactly, at least, and at most.

Probability in Dice Rolling Experiments

Probability helps us measure how likely something is to happen. In dice rolling experiments, probability is based on counting outcomes carefully and comparing the number of outcomes we want to the total number of possible outcomes.

In this lesson, you will learn how to work with one die and two dice, how to list the sample space, and how to find probabilities for sums, doubles, and other conditions such as “greater than” or “less than.”

1. Basic idea of probability

For an experiment with equally likely outcomes, probability is found using the classical formula:

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

A standard die has 6 faces labeled 1 through 6. If the die is fair, each result is equally likely.

So, when rolling one die:

• The sample space is \(\{1,2,3,4,5,6\}\).
• There are 6 possible outcomes.

For example, the probability of rolling a 4 is:

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

The probability of rolling an even number is:

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

because the even outcomes are \(2,4,6\).

2. Rolling two dice

When two dice are rolled, each die can show 6 possible numbers. To count the total number of outcomes, multiply:

$$6 \times 6 = 36$$

So, there are 36 equally likely outcomes when rolling two fair dice.

Each outcome can be written as an ordered pair:

$$ (\text{first die},\text{second die}) $$

For example, \((2,5)\) means the first die shows 2 and the second die shows 5.

Notice that \((2,5)\) and \((5,2)\) are different outcomes, because the dice are being tracked separately.

3. The 36-outcome sample space

Here is the full sample space for rolling two dice:

• First die = 1: \((1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\)
• First die = 2: \((2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\)
• First die = 3: \((3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\)
• First die = 4: \((4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\)
• First die = 5: \((5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\)
• First die = 6: \((6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\)

You can also imagine this as a 6 by 6 grid. This grid is very useful for finding probabilities of sums, doubles, and comparisons.

4. Finding probabilities with two dice

For two-dice experiments, the probability formula becomes:

$$P(\text{event}) = \frac{\text{number of ordered pairs that satisfy the event}}{36}$$

The most important step is to count favorable outcomes correctly.

5. Probabilities of sums

The smallest possible sum is:

$$1+1=2$$

and the largest possible sum is:

$$6+6=12$$

But not all sums are equally likely. Some sums can be made in more ways than others.

Here are the possible sums and how many outcomes give each sum:

• Sum 2: 1 way
• Sum 3: 2 ways
• Sum 4: 3 ways
• Sum 5: 4 ways
• Sum 6: 5 ways
• Sum 7: 6 ways
• Sum 8: 5 ways
• Sum 9: 4 ways
• Sum 10: 3 ways
• Sum 11: 2 ways
• Sum 12: 1 way

This pattern rises to 7 and then falls. That is why 7 is the most likely sum.

6. Probabilities of doubles

A double happens when both dice show the same number.

The doubles are:

$$ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) $$

There are 6 doubles out of 36 total outcomes, so:

$$P(\text{double}) = \frac{6}{36} = \frac{1}{6}$$

7. Probabilities involving inequalities

Sometimes the event involves a comparison, such as:

• the sum is greater than 9
• the first die is less than the second die
• the sum is at most 5

For these questions, it helps to list outcomes carefully or use the 6 by 6 grid.

Words like these are important:

• greater than 9 means \(>9\), so possible sums are 10, 11, 12
• less than 9 means \(<9\)
• at most 5 means \(\le 5\)
• at least 5 means \(\ge 5\)

Worked Example 1: Probability of a specific sum

Question: Two fair dice are rolled. What is the probability that the sum is 8?

Step 1: List the outcomes that give a sum of 8.

$$ (2,6),(3,5),(4,4),(5,3),(6,2) $$

There are 5 favorable outcomes.

Step 2: Divide by the total number of outcomes.

$$P(\text{sum } 8) = \frac{5}{36}$$

Answer: The probability is \(\frac{5}{36}\).

Worked Example 2: Probability of doubles

Question: Two fair dice are rolled. What is the probability of getting a double?

Step 1: List the doubles.

$$ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) $$

There are 6 favorable outcomes.

Step 2: Divide by 36.

$$P(\text{double}) = \frac{6}{36} = \frac{1}{6}$$

Answer: The probability is \(\frac{1}{6}\).

Worked Example 3: Probability with an inequality

Question: Two fair dice are rolled. What is the probability that the sum is greater than 9?

Step 1: “Greater than 9” means the sum can be 10, 11, or 12.

Step 2: Count outcomes for each sum.

• Sum 10: \((4,6),(5,5),(6,4)\) → 3 outcomes
• Sum 11: \((5,6),(6,5)\) → 2 outcomes
• Sum 12: \((6,6)\) → 1 outcome

Total favorable outcomes:

$$3+2+1=6$$

Step 3: Divide by 36.

$$P(\text{sum } > 9) = \frac{6}{36} = \frac{1}{6}$$

Answer: The probability is \(\frac{1}{6}\).

Worked Example 4: Comparing the two dice

Question: Two fair dice are rolled. What is the probability that the first die shows a number less than the second die?

Step 1: Count all outcomes where the first number is smaller than the second.

These are:

• Starting with 1: \((1,2),(1,3),(1,4),(1,5),(1,6)\) → 5 outcomes
• Starting with 2: \((2,3),(2,4),(2,5),(2,6)\) → 4 outcomes
• Starting with 3: \((3,4),(3,5),(3,6)\) → 3 outcomes
• Starting with 4: \((4,5),(4,6)\) → 2 outcomes
• Starting with 5: \((5,6)\) → 1 outcome

Total favorable outcomes:

$$5+4+3+2+1=15$$

Step 2: Divide by 36.

$$P(\text{first die} < \text{second die}) = \frac{15}{36} = \frac{5}{12}$$

Answer: The probability is \(\frac{5}{12}\).

8. Useful strategies for dice probability

1. Find the total number of outcomes first. For two dice, this is 36.
2. Decide exactly what the event means. For example, “sum greater than 8” means 9, 10, 11, or 12.
3. List outcomes carefully. This helps avoid missing possibilities.
4. Remember ordered pairs. \((2,5)\) and \((5,2)\) count as different outcomes.
5. Simplify your fraction if possible.

9. Common mistakes to avoid

• Thinking all sums are equally likely. They are not. For example, a sum of 7 is more likely than a sum of 2.
• Forgetting order matters in the sample space. \((1,6)\) and \((6,1)\) are two different outcomes.
• Using the wrong total. For two dice, the total number of outcomes is 36, not 12.
• Missing outcomes when listing them. Work systematically.

10. Quick practice ideas

Try these on your own:

• What is the probability of a sum of 6?
• What is the probability of a sum less than 5?
• What is the probability of not getting a double?
• What is the probability that the second die is greater than 4?

11. Lesson summary

In dice probability, you compare the number of favorable outcomes to the total number of possible outcomes. For one die, there are 6 outcomes. For two dice, there are 36 ordered outcomes.

To solve problems, list the sample space or count outcomes carefully. This is especially important for sums, doubles, and inequality questions. Once you know how many outcomes satisfy the event, use

$$P(\text{event}) = \frac{\text{favorable outcomes}}{36}$$

for two dice.

With practice, you will get faster at recognizing patterns such as the number of ways to make each sum and the 6 possible doubles.

Probability in Standard Deck Playing Cards

Probability helps us measure how likely something is to happen. A standard deck of playing cards is a great way to study probability because the deck has a fixed number of cards and clear groups, such as suits, colors, and ranks.

In this lesson, you will learn how to find the probability of drawing certain cards from a well-shuffled 52-card deck. You will also learn how to work with events involving suits, colors, face cards, and number cards.

1. The Standard Deck

A standard deck has 52 cards. It is made of:

• 4 suits: hearts, diamonds, clubs, spades
• 2 colors: red and black
• 13 ranks in each suit: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King

The red suits are hearts and diamonds. The black suits are clubs and spades.

Since there are 4 suits and 13 cards in each suit, the total number of cards is:

$$4 \times 13 = 52$$

2. Basic Probability Formula

For equally likely outcomes, probability is found using:

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

When drawing one card from a full deck, the total number of possible outcomes is usually 52.

3. Important Card Counts to Know

These facts are very helpful when solving card probability problems:

• Cards in one suit: 13
• Red cards: 26
• Black cards: 26
• Face cards (Jack, Queen, King): 12
• Aces: 4
• Kings: 4
• Queens: 4
• Jacks: 4
• Cards of any specific rank: 4

Why are there 12 face cards? There are 3 face cards in each suit:

$$4 \times 3 = 12$$

4. Probability of Drawing One Card

If you want the probability of drawing one specific type of card, count how many cards match that description and divide by 52.

For example:

• Probability of drawing a heart: \(\frac{13}{52} = \frac{1}{4}\)
• Probability of drawing a red card: \(\frac{26}{52} = \frac{1}{2}\)
• Probability of drawing a queen: \(\frac{4}{52} = \frac{1}{13}\)

5. Probability of “Or” Events

Sometimes an event can happen in more than one way. For example, you may want the probability of drawing a heart or a king.

When using “or,” be careful not to count a card twice. If one card belongs to both groups, subtract it once.

The formula is:

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

For cards, this often means:

• Count cards in group A
• Count cards in group B
• Subtract any overlap

6. Probability of “And” Events in One Draw

In one draw, “and” usually means one card must satisfy two conditions at the same time.

For example, a card can be red and a queen. The cards that match are the queen of hearts and the queen of diamonds, so there are 2 such cards.

Then:

$$P(\text{red and queen}) = \frac{2}{52} = \frac{1}{26}$$

7. Complement Rule

Sometimes it is easier to find the probability that an event does not happen first.

The complement rule is:

$$P(\text{not }A) = 1 - P(A)$$

For example, instead of counting the probability of drawing a card that is not a face card, you can do:

$$P(\text{not face card}) = 1 - P(\text{face card}) = 1 - \frac{12}{52} = \frac{40}{52} = \frac{10}{13}$$

8. Worked Examples

Example 1: Probability of drawing a spade

Question: What is the probability of drawing a spade from a standard deck?

Step 1: Count favorable outcomes. There are 13 spades.

Step 2: Count total outcomes. There are 52 cards in the deck.

Step 3: Use the formula.

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

Answer: The probability is \(\frac{1}{4}\).

Example 2: Probability of drawing a face card

Question: What is the probability of drawing a face card?

Step 1: Face cards are Jack, Queen, and King.

Step 2: There are 3 face cards in each of the 4 suits.

$$3 \times 4 = 12$$

Step 3: Write the probability.

$$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$$

Answer: The probability is \(\frac{3}{13}\).

Example 3: Probability of drawing a red king

Question: What is the probability of drawing a card that is red and a king?

Step 1: Red suits are hearts and diamonds.

Step 2: The red kings are the king of hearts and the king of diamonds.

So there are 2 favorable cards.

Step 3: Use the probability formula.

$$P(\text{red king}) = \frac{2}{52} = \frac{1}{26}$$

Answer: The probability is \(\frac{1}{26}\).

Example 4: Probability of drawing a club or an ace

Question: What is the probability of drawing a club or an ace?

Step 1: Count clubs. There are 13 clubs.

Step 2: Count aces. There are 4 aces.

Step 3: Find overlap. The ace of clubs was counted in both groups, so subtract 1.

Step 4: Count total favorable outcomes.

$$13 + 4 - 1 = 16$$

Step 5: Write the probability.

$$P(\text{club or ace}) = \frac{16}{52} = \frac{4}{13}$$

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

9. How to Approach Card Probability Questions

When solving a card probability problem, follow these steps:

1. Identify the total number of possible outcomes.
2. Count how many cards match the event.
3. Check whether the event uses and or or.
4. If needed, subtract overlap so cards are not counted twice.
5. Simplify the fraction if possible.

10. Common Mistakes to Avoid

• Forgetting the deck has 52 cards when drawing one card.
• Mixing up suits and colors. A suit is hearts, diamonds, clubs, or spades. A color is red or black.
• Counting overlap twice in “or” problems.
• Using the wrong number of face cards. There are 12 face cards, not 16.
• Not simplifying fractions when possible.

11. Quick Practice Ideas

• What is the probability of drawing a diamond?
• What is the probability of drawing a black card?
• What is the probability of drawing a jack or a heart?
• What is the probability of drawing a card that is not red?

12. Summary

A standard deck has 52 cards split into 4 suits and 2 colors. To find probability, divide the number of favorable outcomes by 52 when drawing one card.

For card problems, it is important to know common card counts, such as 13 cards in each suit, 26 red cards, and 12 face cards. For “or” events, subtract overlap. For “and” events in one draw, count cards that meet both conditions at the same time.

With these ideas, you can calculate the probability of drawing cards based on suit, color, rank, or combinations of these features.

Geometric Probability Models use shapes and area to find the chance that a randomly chosen point lands in a certain region.

Instead of counting outcomes like cards or dice, we compare areas. If a point is equally likely to land anywhere in a figure, then the probability it lands in a shaded part is the fraction of the total area that is shaded.

The main idea is:

$$P(\text{shaded region}) = \frac{\text{area of shaded region}}{\text{area of whole region}}$$

This works only when every point in the larger figure is equally likely to be chosen.

When do we use geometric probability?

• When a point is chosen at random inside a rectangle, circle, triangle, or other figure
• When a dart lands randomly on a board
• When an object lands somewhere along a line segment or inside a region

Main Teaching Point 1: Probability as a ratio of areas

In classical probability, we often use:

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

In geometric probability, the “outcomes” are locations in a figure. Since there are infinitely many points, we cannot count them one by one. Instead, we use length or area.

For 2-dimensional figures, we usually use area:

$$P = \frac{\text{favorable area}}{\text{total area}}$$

If the figure is a line segment, we use length:

$$P = \frac{\text{favorable length}}{\text{total length}}$$

Main Teaching Point 2: The regions must match the same unit

When finding probability, both the numerator and denominator must be measured in the same way.

• Use area over area
• Use length over length

Do not mix area with perimeter, or length with area.

Main Teaching Point 3: Useful area formulas

You will often need these formulas:

• Rectangle: $$A = lw$$
• Square: $$A = s^2$$
• Triangle: $$A = \frac{1}{2}bh$$
• Circle: $$A = \pi r^2$$

Sometimes the shaded region is found by subtraction:

$$\text{shaded area} = \text{whole area} - \text{unshaded area}$$

Main Teaching Point 4: Steps for solving geometric probability problems

1. Identify the whole region.
2. Identify the desired region (usually shaded or described in the problem).
3. Find both areas using the correct formulas.
4. Write the probability as a fraction:

$$P = \frac{\text{desired area}}{\text{total area}}$$

5. Simplify the fraction or convert to a decimal or percent if needed.

Worked Example 1: Simple rectangle

A rectangle has length 10 units and width 6 units. Inside it, a shaded rectangle has length 4 units and width 3 units. A point is chosen at random from the large rectangle. What is the probability that the point lands in the shaded rectangle?

Step 1: Find the area of the whole rectangle.

$$A_{\text{whole}} = 10 \cdot 6 = 60$$

Step 2: Find the area of the shaded rectangle.

$$A_{\text{shaded}} = 4 \cdot 3 = 12$$

Step 3: Form the probability ratio.

$$P = \frac{12}{60} = \frac{1}{5}$$

Answer: The probability is \(\frac{1}{5}\), or 0.2, or 20%.

Worked Example 2: Circle inside a square

A square has side length 8 cm. Inside the square is a circle of radius 4 cm that just fits inside the square. A point is chosen at random inside the square. What is the probability that the point lands inside the circle?

Step 1: Find the area of the square.

$$A_{\text{square}} = 8^2 = 64$$

Step 2: Find the area of the circle.

$$A_{\text{circle}} = \pi(4)^2 = 16\pi$$

Step 3: Write the probability.

$$P = \frac{16\pi}{64} = \frac{\pi}{4}$$

Using a decimal:

$$\frac{\pi}{4} \approx 0.785$$

Answer: The probability is \(\frac{\pi}{4}\), or about 0.785, which is about 78.5%.

Worked Example 3: Shaded region found by subtraction

A large circle has radius 10 units. Inside it is a smaller unshaded circle with radius 6 units. A point is chosen at random inside the large circle. What is the probability that the point lands in the shaded ring between the two circles?

Step 1: Find the area of the large circle.

$$A_{\text{large}} = \pi(10)^2 = 100\pi$$

Step 2: Find the area of the small circle.

$$A_{\text{small}} = \pi(6)^2 = 36\pi$$

Step 3: Find the shaded area.

$$A_{\text{shaded}} = 100\pi - 36\pi = 64\pi$$

Step 4: Find the probability.

$$P = \frac{64\pi}{100\pi} = \frac{16}{25}$$

Answer: The probability is \(\frac{16}{25}\), or 0.64, or 64%.

Worked Example 4: Probability in a triangle

A triangle has base 12 units and height 10 units. Inside it, a shaded triangle has base 6 units and height 4 units. A point is chosen at random inside the large triangle. What is the probability that the point lands in the shaded triangle?

Step 1: Find the area of the large triangle.

$$A_{\text{large}} = \frac{1}{2}(12)(10) = 60$$

Step 2: Find the area of the shaded triangle.

$$A_{\text{shaded}} = \frac{1}{2}(6)(4) = 12$$

Step 3: Find the probability.

$$P = \frac{12}{60} = \frac{1}{5}$$

Answer: The probability is \(\frac{1}{5}\), or 20%.

Common Mistakes to Avoid

• Using side lengths instead of area: In 2D figures, probability usually depends on area, not just one measurement.
• Forgetting subtraction: If the shaded region is the part left over, subtract the inner area from the outer area.
• Using the wrong formula: Make sure you know whether the figure is a rectangle, triangle, or circle.
• Not simplifying: Reduce fractions when possible.

Quick Check Questions

1. A square has side length 5. A shaded rectangle inside it has area 10. What is the probability a random point lands in the shaded region?
2. A circle of radius 3 is inside a square of side length 6. What is the probability a random point in the square lands in the circle?
3. A large rectangle has area 40. A small unshaded rectangle inside it has area 8. What is the probability a random point lands in the shaded part?

Answers to Quick Check

1. Square area: \(5^2 = 25\). Probability: \(\frac{10}{25} = \frac{2}{5}\).
2. Circle area: \(\pi(3)^2 = 9\pi\). Square area: \(6^2 = 36\). Probability: \(\frac{9\pi}{36} = \frac{\pi}{4}\).
3. Shaded area: \(40 - 8 = 32\). Probability: \(\frac{32}{40} = \frac{4}{5}\).

Summary

Geometric probability is used when outcomes are points in a figure instead of countable objects. If every point in the larger region is equally likely, then the probability of landing in a certain part is the area of that part divided by the total area.

Always identify the whole figure, find the desired region, calculate the areas carefully, and then write the probability as a ratio. With practice, these problems become a straightforward use of area formulas and fractions.

Expected Value and Game Fairness

When people play games of chance, they often ask, “Is this game fair?” A game may look fair because everyone follows the same rules, but in mathematics, fairness has a more specific meaning.

To decide whether a game is fair, we use a probability idea called expected value. Expected value helps us measure the average result a player can expect over many plays of the game.

This lesson will show you how to calculate expected value and how to use it to decide whether a game gives an equal advantage to the players.

1. What is expected value?

The expected value of a game is the average amount of money won or lost per play if the game is repeated many times.

It does not mean you will get that exact amount in one play. Instead, it tells you what tends to happen in the long run.

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

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

If there are only a few outcomes, you can write it as:

$$ E = x_1P(x_1) + x_2P(x_2) + x_3P(x_3) + \cdots $$

Here:

• Outcome means the amount won or lost.
• Probability means the chance that outcome happens.

2. Positive and negative values

When working with games, money won is written as a positive number, and money lost is written as a negative number.

For example:

• Winning $5 is written as \(+5\).
• Losing $2 is written as \(-2\).

This is important because expected value depends on both gains and losses.

3. What makes a game fair?

A game is called fair if the expected value is 0. That means, on average, a player neither gains nor loses money over many plays.

A game is:

• Fair if expected value \(= 0\)
• Favorable to the player if expected value \(> 0\)
• Favorable to the organizer if expected value \(< 0\)

So, to test fairness, we calculate the expected value from the player’s point of view.

4. Steps for solving expected value problems

1. List all possible outcomes.
2. Find the probability of each outcome.
3. Write each money result as positive or negative.
4. Multiply each outcome by its probability.
5. Add the products.
6. Use the result to decide if the game is fair.

Worked Example 1: A simple spinner game

A spinner has 4 equal sections. If it lands on red, a player wins $8. If it lands on any other color, the player loses $2. Is the game fair?

Step 1: Identify the outcomes.

• Win $8 with probability \(\frac{1}{4}\)
• Lose $2 with probability \(\frac{3}{4}\)

Step 2: Use the expected value formula.

$$ E = 8\left(\frac{1}{4}\right) + (-2)\left(\frac{3}{4}\right) $$ $$ E = 2 + \left(-\frac{6}{4}\right) $$ $$ E = 2 - 1.5 = 0.5 $$

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

Conclusion: The game is not fair. It is favorable to the player because the player gains an average of 50 cents per play.

Worked Example 2: A die game with an entry fee

A player pays $3 to roll a fair six-sided die.

• If the player rolls a 6, they receive $10.
• If the player rolls 1 through 5, they receive nothing.

Is the game fair?

The player first pays $3, so we must look at the net result.

• If the player rolls a 6, they receive $10 but paid $3, so net gain is \(+7\).
• If the player rolls 1 through 5, they receive $0 but paid $3, so net result is \(-3\).

The probabilities are:

• Rolling a 6: \(\frac{1}{6}\)
• Rolling 1 through 5: \(\frac{5}{6}\)

Now calculate expected value:

$$ E = 7\left(\frac{1}{6}\right) + (-3)\left(\frac{5}{6}\right) $$ $$ E = \frac{7}{6} - \frac{15}{6} $$ $$ E = -\frac{8}{6} = -\frac{4}{3} $$

So the expected value is \(-\$1.33\) approximately.

Conclusion: The game is not fair. It is favorable to the organizer, because the player loses about $1.33 on average per play.

5. Fair price and fair entry fee

Sometimes a question asks what entry fee would make a game fair. In that case, we set the expected value equal to 0 and solve.

Worked Example 3: Finding a fair entry fee

A bag contains 5 marbles: 1 gold marble and 4 blue marbles. A player draws one marble at random.

• If the player draws the gold marble, they win $12.
• If the player draws a blue marble, they win nothing.

What entry fee would make the game fair?

Let the entry fee be \(x\) dollars.

Then the player’s net result is:

• Gold marble: \(12 - x\)
• Blue marble: \(-x\)

The probabilities are:

• Gold marble: \(\frac{1}{5}\)
• Blue marble: \(\frac{4}{5}\)

For a fair game, expected value must equal 0:

$$ (12 - x)\left(\frac{1}{5}\right) + (-x)\left(\frac{4}{5}\right) = 0 $$

Multiply through by 5:

$$ 12 - x - 4x = 0 $$ $$ 12 - 5x = 0 $$ $$ 5x = 12 $$ $$ x = 2.4 $$

The fair entry fee is $2.40.

If the organizer charges more than $2.40, the game favors the organizer. If the organizer charges less than $2.40, the game favors the player.

6. Expected value in words

It can help to describe expected value in everyday language:

• An expected value of \(0\) means break-even in the long run.
• A positive expected value means average gain in the long run.
• A negative expected value means average loss in the long run.

This does not guarantee what happens in one game. A player can still win a game with negative expected value or lose a game with positive expected value. Expected value only describes the average over many repeated plays.

Worked Example 4: A game with more than two outcomes

A player rolls a fair six-sided die.

• If the result is 1 or 2, the player wins $4.
• If the result is 3, 4, or 5, the player loses $1.
• If the result is 6, the player loses $5.

Find the expected value and decide whether the game is fair.

Step 1: Write outcomes and probabilities.

• Win $4 with probability \(\frac{2}{6} = \frac{1}{3}\)
• Lose $1 with probability \(\frac{3}{6} = \frac{1}{2}\)
• Lose $5 with probability \(\frac{1}{6}\)

Step 2: Calculate expected value.

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

Use a common denominator of 6:

$$ E = \frac{8}{6} - \frac{3}{6} - \frac{5}{6} $$ $$ E = \frac{0}{6} = 0 $$

Conclusion: The expected value is 0, so the game is fair.

7. Common mistakes to avoid

• Forgetting losses are negative. If a player loses money, use a negative number.
• Ignoring the entry fee. Always use the net gain or net loss after the fee is paid.
• Using wrong probabilities. Make sure all probabilities add up to 1.
• Thinking expected value predicts one play. It only describes the average over many plays.

8. Quick check questions

Try these on your own:

1. A coin is flipped. If it lands heads, you win $3. If it lands tails, you lose $1. What is the expected value?
2. You pay $2 to draw one card from 10 cards. One card wins $9, and the other 9 cards win nothing. Is the game fair?
3. A game has expected value \(-0.75\). Who does the game favor?

Answers:

1. \(E = 3\left(\frac{1}{2}\right) + (-1)\left(\frac{1}{2}\right) = 1\). The expected value is $1.
2. Net outcomes: \(+7\) with probability \(\frac{1}{10}\), and \(-2\) with probability \(\frac{9}{10}\).
   $$E = 7\left(\frac{1}{10}\right) + (-2)\left(\frac{9}{10}\right) = \frac{7}{10} - \frac{18}{10} = -\frac{11}{10} = -1.1$$ The game is not fair; it favors the organizer.
3. The game favors the organizer because the player loses 75 cents on average per play.

9. Summary

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

A game is fair when the expected value is 0. If the expected value is positive, the game favors the player. If it is negative, the game favors the organizer.

When analyzing game fairness, always pay attention to probabilities, gains, losses, and entry fees. With these steps, you can use mathematics to decide whether a game of chance is truly fair.

Sampling With and Without Replacement

In probability, we often choose objects from a group and ask, “What is the chance of getting a certain result?”

When we choose more than one object, an important question is whether the first object is put back before the next choice. This changes the probabilities.

This idea is called sampling with replacement and sampling without replacement.

Understanding the difference helps you solve many probability problems involving cards, marbles, counters, names, and more.

1. Key idea: What does “replacement” mean?

Sampling with replacement means you choose an item, record it, and then put it back before choosing again.

Because the item is returned, the total number of items in the group stays the same. The probabilities for the next draw also stay the same.

Sampling without replacement means you choose an item and do not put it back before the next choice.

Because the item is not returned, the total number of items decreases. This means the probabilities for the next draw can change.

2. Why this matters in probability

Suppose a bag has 5 red counters and 3 blue counters, for a total of 8 counters.

The probability of drawing a red counter first is

$$P(\text{red first})=\frac{5}{8}$$

Now think about the second draw.

• With replacement: the first counter is returned, so there are still 5 red out of 8 total. The probability of red on the second draw is still \(\frac{5}{8}\).
• Without replacement: the first counter is not returned, so the numbers in the bag change. The probability on the second draw depends on what happened first.

This is the big difference: without replacement, earlier draws affect later draws.

3. Independent and dependent events

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

Two events are dependent if one event does change the probability of the other.

• With replacement usually gives independent events, because the group goes back to how it started.
• Without replacement gives dependent events, because the first result changes the group.

4. Finding probabilities of two draws

When we want the probability of two events both happening, we multiply along the path:

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

Here, \(P(B\mid A)\) means “the probability of \(B\), given that \(A\) has already happened.”

For with replacement, the second probability often stays the same as the first because nothing changed.

For without replacement, the second probability often changes because one item has been removed.

5. Worked Example 1: With replacement

A bag contains 4 green marbles and 6 yellow marbles. Two marbles are drawn with replacement. What is the probability that both marbles are green?

Step 1: Find the probability of green on the first draw.

There are 4 green marbles out of 10 total.

$$P(\text{green first})=\frac{4}{10}=\frac{2}{5}$$

Step 2: Find the probability of green on the second draw.

Because the marble is replaced, the bag is unchanged.

$$P(\text{green second})=\frac{4}{10}=\frac{2}{5}$$

Step 3: Multiply.

$$P(\text{both green})=\frac{2}{5}\times\frac{2}{5}=\frac{4}{25}$$

Answer: The probability is \(\frac{4}{25}\).

6. Worked Example 2: Without replacement

A bag contains 4 green marbles and 6 yellow marbles. Two marbles are drawn without replacement. What is the probability that both marbles are green?

Step 1: First draw green.

$$P(\text{green first})=\frac{4}{10}=\frac{2}{5}$$

Step 2: Second draw green.

Now one green marble has already been removed. So there are 3 green marbles left out of 9 total.

$$P(\text{green second}\mid\text{green first})=\frac{3}{9}=\frac{1}{3}$$

Step 3: Multiply.

$$P(\text{both green})=\frac{2}{5}\times\frac{1}{3}=\frac{2}{15}$$

Answer: The probability is \(\frac{2}{15}\).

Compare the two examples:

• With replacement: \(\frac{4}{25}\)
• Without replacement: \(\frac{2}{15}\)

These are different because in the second case, the first draw changes what is left in the bag.

7. Worked Example 3: Different colors in order

A box contains 3 red pens and 2 blue pens. Two pens are chosen without replacement. What is the probability of choosing a red pen first and then a blue pen?

Step 1: Probability of red first.

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

Step 2: Probability of blue second, given red first.

After taking one red pen, 4 pens remain: 2 red and 2 blue.

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

Step 3: Multiply.

$$P(\text{red then blue})=\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$$

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

Notice that this question asks for a specific order: red first, then blue.

If a problem does not care about order, you may need to add more than one case.

8. Worked Example 4: Probability without caring about order

A jar contains 5 black beads and 3 white beads. Two beads are chosen without replacement. What is the probability of getting one black bead and one white bead?

This can happen in two ways:

1. Black first, then white
2. White first, then black

Case 1: Black then white

$$P(B\text{ then }W)=\frac{5}{8}\times\frac{3}{7}=\frac{15}{56}$$

Case 2: White then black

$$P(W\text{ then }B)=\frac{3}{8}\times\frac{5}{7}=\frac{15}{56}$$

Add the cases.

$$P(\text{one black and one white})=\frac{15}{56}+\frac{15}{56}=\frac{30}{56}=\frac{15}{28}$$

Answer: The probability is \(\frac{15}{28}\).

9. Using tree diagrams

A tree diagram is a useful way to organize probability problems with more than one step.

For example, if a bag has 2 red and 1 blue counter, and two counters are drawn without replacement, a tree diagram helps show how the second probabilities change.

You would begin with branches for the first draw:

• Red: \(\frac{2}{3}\)
• Blue: \(\frac{1}{3}\)

Then from each branch, draw the possible second outcomes using the new totals.

After a red is drawn first, there are 1 red and 1 blue left, so the second-draw probabilities are different from the start.

Tree diagrams are especially helpful when:

• there are two or more draws,
• the problem includes replacement or no replacement,
• you need to keep track of order.

10. Quick comparison: with vs without replacement

• With replacement: put the item back
• Without replacement: do not put the item back
• With replacement: total number stays the same
• Without replacement: total number goes down
• With replacement: later probabilities usually stay the same
• Without replacement: later probabilities usually change
• With replacement: events are usually independent
• Without replacement: events are dependent

11. Common mistakes to avoid

• Forgetting to change the total when there is no replacement. If one item is removed, the total number of items becomes one less.
• Forgetting to change the number of wanted items. For example, if you already drew one red object, there may be fewer red objects left.
• Mixing up “and” and “or”. “And” usually means multiply. “Or” often means add different cases.
• Ignoring order when the problem cares about it, or forgetting to include both orders when the problem does not care.

12. A strategy for solving these problems

1. Read carefully: Is it with replacement or without replacement?
2. Find the probability of the first event.
3. Update the numbers if there is no replacement.
4. Find the probability of the second event.
5. Multiply for events happening together.
6. If the question does not care about order, add all valid cases.

13. Practice check

Try these on your own:

• A bag has 7 red counters and 3 blue counters. Two counters are drawn with replacement. What is the probability both are blue?
• A bag has 7 red counters and 3 blue counters. Two counters are drawn without replacement. What is the probability both are blue?
• A box has 4 white balls and 5 black balls. Two balls are drawn without replacement. What is the probability of one white and one black?

Answers:

• With replacement: $$\frac{3}{10}\times\frac{3}{10}=\frac{9}{100}$$
• Without replacement: $$\frac{3}{10}\times\frac{2}{9}=\frac{1}{15}$$
• One white and one black: $$\frac{4}{9}\times\frac{5}{8}+\frac{5}{9}\times\frac{4}{8}=\frac{40}{72}=\frac{5}{9}$$

14. Summary

Sampling with replacement means the chosen item goes back, so the total number of items does not change and the probabilities usually stay the same.

Sampling without replacement means the chosen item is not returned, so the totals change and later probabilities also change.

To solve these problems, pay close attention to whether the group changes after each draw, then multiply probabilities along each path and add cases when needed.