Chapter 6 Probability
6.2 Conditional Probability and Bayes’ Theorem
Learning Objectives
By the end of this section, you will be able to:
- Calculate conditional probability
- Use Bayes’ theorem to determine the probability of an event
Often it is required to compute the probability of an event given that another event has occurred. We call that conditional probability.
Conditional Probability
The probability the event B occurs, given that event A has happened, is represented as [latex]P(B|A)[/latex]. This is read as “the probability of B given A.”
Example 1
What is the probability that two cards drawn at random from a deck of playing cards will both be aces?
It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\frac{4}{52} \cdot \frac{4}{52} = \frac{1}{169}[/latex]. This would be incorrect, however, because the two events are not independent. The first card is not replaced before the second card is drawn. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.
Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case the “condition” is that the first card is an ace. Symbolically, we write this as:
[latex]P(\text{ace on the second draw} | \text{ace on the first draw})[/latex]
The vertical bar “|” is read as “given,” so the above expression is short for “The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw.” What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]\frac{3}{51} = \frac{1}{17}[/latex].
Thus, the probability of both cards being aces is [latex]\frac{4}{52} \cdot \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}[/latex].
Example 2
Find the probability that a die rolled shows a 6 given that a flipped coin shows a head.
These are two independent events, so the probability of the die rolling a 6 is [latex]\frac{1}{6}[/latex] regardless of the result of the coin flip.
Example 3
The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year and the color of their car.
[latex]\begin{array} {|c|c|c|c|} \hline \text{}& \text{Speeding} &\text{No speeding} & \text{Total}\\ &\text{ticket}& \text{ticket}\\ \hline \text{Red car} & 15 & 135 & 150\\ \hline \text{Not red car} & 45 & 470 & 515\\ \hline \text{Total} & 60 & 605 & 665\\ \hline \end{array}[/latex]
Find the probability that a randomly chosen person:
a) Has a speeding ticket given they have a red car
Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so [la