# 2.3: Conditional Probability

**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:

\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.

*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 \begin{align*}(\clubsuit)\end{align*}

*Solution:*

**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%.

*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

\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%.

*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.

\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%.

*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 \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?

*Solution:*

**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%.

**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 \begin{align*}(_nC_r)\end{align*} of objects \begin{align*}(r)\end{align*} without regard to order and without repetition selected from a distinct number of objects \begin{align*}(n)\end{align*}.

- 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 \begin{align*}p\end{align*} different ways and another independent event can be chosen in \begin{align*}q\end{align*} different ways, the probability of the 2 events occurring is \begin{align*}p \times q\end{align*}.

- Permutations
- The number of possible arrangements \begin{align*}({_n}P_r)\end{align*} in an ordered set of objects, where \begin{align*}n =\end{align*} the number of objects and \begin{align*}r =\end{align*} 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.