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# Conditional Probability

## P(B|A) = P(A and B)/ P(A)

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Conditional Probability

You've just done some data collection to determine the popularity of courses at your high school. Your calculations show that 75% of the students take Geometry and 15% of the students take both Chemistry and Geometry. How would you find the probability that a student who is taking Chemistry is also taking Geometry?

### Watch This

First watch this video to learn about conditional probability.

Then watch this video to see some examples.

Watch this video for more help.

### Guidance

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:

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

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:

$P(B | A) &= \frac{P(A \cap B)}{P(A)}\\P(A \cap B) &= P(A) \times P(B | A)$

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:

$P(\text{second event} | \text{first event})=\frac{P(\text{first event and second event})}{P(\text{first event})}$

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

#### Example A

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 $(\clubsuit)$ . 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

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

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

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

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

#### Example B

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

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

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

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

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

#### Example C

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 $\frac{1}{2}$ . Based on past runs, the probability of turning right the second time is $\frac{2}{3}$ . 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.

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

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

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

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

### Guided Practice

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.

$P(\text{physics}) &= 0.90\\P(\text{physics and statistics}) &= 0.35$

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

$P(\text{statistics} | \text{physics}) &= \frac{P(\text{physics and statistics})}{P(\text{physics})}\\P(\text{statistics} | \text{physics}) &= \frac{0.35}{0.90}\\P(\text{statistics} | \text{physics}) &= 0.388\\P(\text{statistics} | \text{physics}) &= 39\%$

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

### Practice

1. 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. 2 fair dice are rolled. What is the probability that the sum is even given that the first die rolled is a 5?
3. 2 fair dice are rolled. What is the probability that the sum is odd given that the first die rolled is a 5?
4. 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?
5. 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?
6. 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?
7. 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?
8. 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?
9. 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?
10. For question 9, what is the probability that you don't have the disease, given that you tested negative?

### Vocabulary Language: English Spanish

conditional probability

conditional probability

The probability of a particular dependent event  given the outcome of the event on which it occurs.
conditional probability formula

conditional probability formula

The conditional probability formula is P(A/B) = P(AUB)/P(B)
Dependent Events

Dependent Events

In probability situations, dependent events are events where one outcome impacts the probability of the other.
Favorable Outcome

Favorable Outcome

A favorable outcome is the outcome that you are looking for in an experiment.
Independent Events

Independent Events

Two events are independent if the occurrence of one event does not impact the probability of the other event.
Multiplication Rule

Multiplication Rule

States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
Mutually Exclusive Events

Mutually Exclusive Events

Mutually exclusive events have no common outcomes.
Sample Space

Sample Space

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