<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Basic Probability and Statistics Go to the latest version.

# 2.3: Conditional Probability

Difficulty Level: At Grade Created by: CK-12

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 18

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?

Solution:

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 19

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?

Solution:

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 20

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?

Solution:

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%.

Example 21

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?

Solution:

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%.

Points to Consider

• How does a permutation differ from a combination?
• How are tree diagrams helpful for determining probabilities?

Vocabulary

Combinations
The number of possible arrangements $(_nC_r)$ of objects $(r)$ without regard to order and without repetition selected from a distinct number of objects $(n)$.
Conditional probability
The probability of a particular dependent event, given the outcome of the event on which it depends.
Factorial function (!)
To multiply a series of consecutive descending natural numbers.
Fundamental Counting Principle
If an event can be chosen in $p$ different ways and another independent event can be chosen in $q$ different ways, the probability of the 2 events occurring is $p \times q$.
Permutations
The number of possible arrangements $({_n}P_r)$ in an ordered set of objects, where $n =$ the number of objects and $r =$ the number of objects selected.
Tree diagrams
A way to show the outcomes of simple probability events, where each outcome is represented as a branch on a tree.

Feb 23, 2012

Jan 27, 2015