### Conditional Probability

What if the probability of a second event is affected by the probability of the first event? This type of probability calculation is known as **conditional probability**.

When working with events that are conditionally probable, you are working with 2 events, where the probability of the second event is conditional on the first event occurring. Say, for example, that you want to know the probability of drawing 2 kings from a deck of cards. As we have previously learned, here is how you would calculate this:

\begin{align*}P(\text{first king}) &= \frac{1}{13}\\
P(\text{second king}) &= \frac{3}{51}\\
P(\text{2 kings}) &= \frac{1}{13} \times \frac{3}{51}\\
P(\text{2 kings}) &= \frac{3}{663}\\
P(\text{2 kings}) &= \frac{1}{221}\\\end{align*}

Now let’s assume you are playing a game where you need to draw 2 kings to win. You draw the first card and get a king. What is the probability of getting a king on the second card? The probability of getting a king on the second card can be thought of as a conditional probability. The formula for calculating conditional probability is given as:

\begin{align*}P(B | A) &= \frac{P(A \cap B)}{P(A)}\\
P(A \cap B) &= P(A) \times P(B | A)\end{align*}

Another way to look at the conditional probability formula is as follows. Assuming the first event has occurred, the probability of the second event occurring is:

\begin{align*}P(\text{second event} | \text{first event})=\frac{P(\text{first event and second event})}{P(\text{first event})}\end{align*}

Let’s work through a few problems using the formula for conditional probability.

#### Using the Formula for Conditional Probability

1. You are playing a game of cards where the winner is determined when a player gets 2 cards of the same suit. You draw a card and get a club \begin{align*}(\clubsuit)\end{align*}. What is the probability that the second card will be a club?

**Step 1:** List what you know.

First event = drawing the first club

Second event = drawing the second club

\begin{align*}P(\text{first club}) &= \frac{13}{52}\\ P(\text{second club}) &= \frac{12}{51}\\ P(\text{club and club}) &= \frac{13}{52} \times \frac{12}{51}\\ P(\text{club and club}) &= \frac{156}{2,652}\\ P(\text{club and club}) &= \frac{1}{17}\end{align*}

**Step 2:** Calculate the probability of choosing a club as the second card when a club is chosen as the first card.

\begin{align*}\text{Probability of drawing the second club} &= \frac{P(\text{club and club})}{P(\text{first club})}\\ P(\text{club} | \text{club}) &= \frac{\frac{1}{17}}{\frac{13}{52}}\\ P(\text{club} | \text{club}) &= \frac{1}{17} \times \frac{52}{13}\\ P(\text{club} | \text{club}) &= \frac{52}{221}\\ P(\text{club} | \text{club}) &= \frac{4}{17}\end{align*}

**Step 3:** Write your conclusion.

Therefore, the probability of selecting a club as the second card when a club is chosen as the first card is 24%.

2. In the next round of the game, the first person to be dealt a black ace wins the game. You get your first card, and it is a queen. What is the probability of obtaining a black ace?

**Step 1:** List what you know.

First event = being dealt the queen

Second event = being dealt the black ace

\begin{align*}P(\text{queen}) &= \frac{4}{52}\\ P(\text{black ace}) &= \frac{2}{51}\\ P(\text{black ace and queen}) &= \frac{4}{52} \times \frac{2}{51}\\ P(\text{black ace and queen}) &= \frac{8}{2,652}\\ P(\text{black ace and queen}) &= \frac{2}{663}\end{align*}

**Step 2:** Calculate the probability of choosing black ace as a second card when a queen is chosen as a first card.

\begin{align*}P(\text{black ace}|\text{queen}) &= \frac{P(\text{black ace and queen})}{P(\text{queen})}\\ P(\text{black ace} | \text{queen}) &= \frac{\frac{2}{663}}{\frac{4}{52}}\\ P(\text{black ace} | \text{queen}) &= \frac{2}{663} \times \frac{52}{4}\\ P(\text{black ace} | \text{queen}) &= \frac{104}{2,652}\\ P(\text{black ace} | \text{queen}) &= \frac{2}{51}\end{align*}

**Step 3:** Write your conclusion.

Therefore, the probability of selecting a black ace as the second card when a queen is chosen as the first card is 3.9%.

#### Calculating Probability

Sandra went out for her daily run. She goes on a path that has alternate routes to give her a variety of choices to make her run more enjoyable. The path has 3 turns where she can go left or right at each turn. The probability of turning right the first time is \begin{align*}\frac{1}{2}\end{align*}. Based on past runs, the probability of turning right the second time is \begin{align*}\frac{2}{3}\end{align*}. Draw a tree diagram to represent the path. What is the probability that she will turn left the second time after turning right the first time?

**Step 1:** List what you know.

\begin{align*}P(\text{right the first time}) &= \frac{1}{2}\\ P(\text{right the second time}) &= \frac{2}{3}\\ P(\text{left the second time}) &= 1 - \frac{2}{3} = \frac{1}{3}\\ P(\text{right the first time and left the second time}) &= \frac{1}{2} \times \frac{1}{3}\\ P(\text{right the first time and left the second time}) &= \frac{1}{6}\end{align*}

**Step 2:** Calculate the probability of choosing left as the second turn when right is chosen as the first turn.

\begin{align*}P(\text{left the second time} | \text{right the first time}) &= \frac{P(\text{right the first time and left the second time})}{P(\text{right the first time})}\\ P(\text{left the second time} | \text{right the first time}) &= \frac{\frac{1}{6}}{\frac{1}{2}}\\ P(\text{left the second time} | \text{right the first time}) &= \frac{1}{6} \times \frac{2}{1}\\ P(\text{left the second time} | \text{right the first time}) &= \frac{2}{6}\\ P(\text{left the second time} | \text{right the first time}) &= \frac{1}{3}\\ P(\text{left the second time} | \text{right the first time}) &= 0.33\overline{3}\\ P(\text{left the second time} | \text{right the first time}) &= 33\%\end{align*}

**Step 3:** Write your conclusion.

Therefore, the probability of choosing left as the second turn when right was chosen as the first turn is 33%.

### Example

#### Example 1

At Bluenose High School, 90% of the students take Physics and 35% of the students take both Physics and Statistics. What is the probability that a student from Bluenose High School who is taking Physics is also taking Statistics?

**Step 1:** List what you know.

\begin{align*}P(\text{physics}) &= 0.90\\ P(\text{physics and statistics}) &= 0.35\end{align*}

**Step 2:** Calculate the probability of choosing Statistics as a second course when Physics is chosen as a first course.

\begin{align*}P(\text{statistics} | \text{physics}) &= \frac{P(\text{physics and statistics})}{P(\text{physics})}\\ P(\text{statistics} | \text{physics}) &= \frac{0.35}{0.90}\\ P(\text{statistics} | \text{physics}) &= 0.388\\ P(\text{statistics} | \text{physics}) &= 39\%\end{align*}

**Step 3:** Write your conclusion.

Therefore, the probability that a student from Bluenose High School who is taking Physics is also taking Statistics is 39%.

### Review

- 2 fair dice are rolled. What is the probability that the sum is even given that the first die that is rolled is a 2?
- 2 fair dice are rolled. What is the probability that the sum is even given that the first die rolled is a 5?
- 2 fair dice are rolled. What is the probability that the sum is odd given that the first die rolled is a 5?
- Steve and Scott are playing a game of cards with a standard deck of playing cards. Steve deals Scott a black king. What is the probability that Scott’s second card will be a red card?
- Sandra and Karen are playing a game of cards with a standard deck of playing cards. Sandra deals Karen a red seven. What is the probability that Karen’s second card will be a black card?
- Donna discusses with her parents the idea that she should get an allowance. She says that in her class, 55% of her classmates receive an allowance for doing chores, and 25% get an allowance for doing chores and are good to their parents. Her mom asks Donna what the probability is that a classmate will be good to his or her parents given that he or she receives an allowance for doing chores. What should Donna's answer be?
- At a local high school, the probability that a student speaks English and French is 15%. The probability that a student speaks French is 45%. What is the probability that a student speaks English, given that the student speaks French?
- At a local high school, the probability that a student takes statistics and art is 10%. The probability that a student takes art is 65%. What is the probability that a student takes statistics, given that the student takes art?
- The test for a disease is accurate 80% of the time, and 2.5% of the population has the disease. What is the probability that you have the disease, given that you tested positive?
- For question 9, what is the probability that you don't have the disease, given that you tested negative?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 2.7.