Often it is required to compute the probability of an event given that another event has occurred.
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]\displaystyle\frac{{4}}{{52}}\times\frac{{4}}{{52}}=\frac{{1}}{{169}}[/latex]. This would be
incorrect, however, because the two events are not independent. 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:
P(ace on second draw | an ace on the first draw).
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]\displaystyle\frac{{3}}{{51}}=\frac{{1}}{{17}}[/latex].
Thus, the probability of both cards being aces is
[latex]\displaystyle\frac{{4}}{{52}}\times\frac{{3}}{{51}}=\frac{{12}}{{2652}}=\frac{{1}}{{221}}[/latex].
Conditional Probability
The probability the event
B occurs, given that event
A has happened, is represented as
P(
B |
A)
This is read as "the probability of
B given
A"
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]\displaystyle\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. Find the probability that a randomly chosen person:
- Has a speeding ticket given they have a red car
- Has a red car given they have a speeding ticket
|
Speeding ticket |
No speeding ticket |
Total |
| Red car |
15 |
135 |
150 |
| Not red car |
45 |
470 |
515 |
| Total |
60 |
605 |
665 |
- 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
P(ticket | red car) =
[latex]\displaystyle\frac{{15}}{{150}}=\frac{{1}}{{110}}={0.1}[/latex]
- Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so
P(red car | ticket) =
[latex]\displaystyle\frac{{15}}{{60}}=\frac{{1}}{{4}}={0.25}[/latex]
Notice from the last example that P(B | A) is
not equal to P(A | B).
These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.
Conditional Probability Formula
If Events
A and
B are not independent, then
P(
A and
B) =
P(
A)
P(
B |
A)
Example 4
If you pull 2 cards out of a deck, what is the probability that both are spades?
The probability that the first card is a spade is
[latex]\displaystyle\frac{{13}}{{52}}[/latex].
The probability that the second card is a spade, given the first was a spade, is
[latex]\displaystyle\frac{{12}}{{51}}[/latex], since there is one less spade in the deck, and one less total cards.
The probability that both cards are spades is
[latex-display]\displaystyle\frac{{13}}{{52}}\times\frac{{12}}{{51}}=\frac{{156}}{{2652}}\approx{0.0588}[/latex-display]
Example 5
If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?
You can satisfy this condition by having Case A or Case B, as follows:
- Case A) you can get the Ace of Diamonds first and then a black card or
- Case B) you can get a black card first and then the Ace of Diamonds.
Let's calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is
[latex]\displaystyle\frac{{1}}{{52}}[/latex].
Now for Case B: the probability that the first card is black is
[latex]\displaystyle\frac{{26}}{{52}}=\frac{{1}}{{2}}[/latex], the same as the probability of Case 1.
Recall that the probability of A or B is
P(A) +
P(B) –
P(A and B). In this problem,
P(A and B) = 0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is [latex]\displaystyle\frac{{1}}{{101}}+\frac{{1}}{{101}}=\frac{{2}}{{101}}[/latex].
Try it Now 1
In your drawer you have 10 pairs of socks, 6 of which are white. If you reach in and randomly grab two pairs of socks, what is the probability that both are white?
Example 6
A home pregnancy test was given to women, then pregnancy was verified through blood tests. The following table shows the home pregnancy test results. Find
- P(not pregnant | positive test result)
- P(positive test result | not pregnant)
|
Positive test |
Negative test |
Total |
| Pregnant |
70 |
4 |
74 |
| Not Pregnant |
5 |
14 |
19 |
| Total |
75 |
18 |
93 |
- Since we know the test result was positive, we're limited to the 75 women in the first column, of which 5 were not pregnant. P(not pregnant | positive test result) = [latex]\displaystyle\frac{{5}}{{75}}\approx{0.067}[/latex]
- Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test. P(positive test result | not pregnant) = [latex]\displaystyle\frac{{5}}{{19}}\approx{0.263}[/latex]
The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.
Try It Now Answer
- [latex]\displaystyle\frac{{6}}{{10}}\times\frac{{5}}{{9}}=\frac{{30}}{{90}}=\frac{{1}}{{3}}[/latex]
David Lippman, Math in Society, "Probability," licensed under a CC BY-SA 3.0 license.
