5.5: Conditional Probabilities
Students will be able to:
- Calculate the conditional probability of an event A given that event B has already occurred.
- Identify instances when the occurance of one event will influence the probably of another event.
- Compute the conditional probability of an event by using a two-way table.
What do you think the probability is that a man is over six feet tall? If you knew that both his parents were tall would you change your estimate of the probability? A conditional probability is a probability that is based on some prior knowledge.
A conditional probability is the probability that an event will occur if some other condition has already occurred. This is denoted by \(P(A | B)\), which is read “the probability of A given B .”
Two cards are drawn from a well shuffled deck of 52 cards without replacement. Find the following probabilities:
- The probability that the second card is a heart given that the first card is a spade.
- The probability that the first card is a face card and the second card an ace.
- The probability that one card is a heart and the other a club.
Solution a
Without replacement means that the first card is set aside before the second card is drawn and we assume the first card is a spade. There are only 51 cards to choose from for the second card. Thirteen of those cards are hearts.
It’s important to notice that the question only asks about the second card.
\[ P(\text{2nd heart | 1st spade}) = \dfrac{13}{51} \nonumber\]
The probability that the second card is a heart given that the first card is a spade is \(\dfrac{13}{51}\).
Solution b
Notice that this time the question asks about both of the cards.
There are 12 face cards out of 52 cards when we draw the first card. We set the first card aside and assume that it is a face card. Then there are four aces out of the 51 remaining cards. We want to draw a face card and an ace so use multiplication.
\[ P(\text{1st face card and 2nd ace}) = \dfrac{12}{52} \cdot \dfrac{4}{51} = \dfrac{48}{2652} \approx 0.018 \nonumber\]
The probability that the first card is a face card and the second card an ace is approximately 0.018 or 1.8%.
Solution c
There are two ways for this to happen. We could get a heart first and a club second or we could get the club first and the heart second.
\[ \begin{align*} P(\text{heart and club}) &= P(\text{heart 1st and club 2nd or club 1st and heart 2nd}) \\[4pt]&= P(\text{heart 1st and club 2nd}) + P(\text{club 1st and heart 2nd}) \\[4pt]&= \dfrac {13}{52} \cdot \dfrac {13}{51} + \dfrac {13}{52} \cdot \dfrac {13}{51} \\[4pt]&\approx 0.127 \end{align*}\]
The probability that one card is a heart and the other a club is approximately 0.127 or 12.7%.
Two fair dice are rolled and the sum of the numbers is observed. What is the probability that the sum is at least nine if it is known that one of the dice shows a five?
Since we are given that one of the dice shows a five this is a conditional probability. List the pairs of dice with one die showing a five. Be careful not to count (5,5) twice.
\[\{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6)\} \nonumber\]
List the pairs from above that have a sum of at least nine.
\[\{(4,5), (5,5), (6,5), (5,4), (5,6)\}\nonumber\]
Solution
There are 11 ways for one die to show a five and five of these ways have a sum of at least nine.
\[ P(\text{sum at least 9 | one die is a 5}) = \dfrac{5}{11} \nonumber\]
The probability that the sum is at least nine if it is known that one of the dice shows a five is \(\dfrac{5}{11}\).
For events A and B, \( P(A \text{ and } B) = P(A) \cdot P(B|A) \)
For events A and B, \( P(B|A) = \dfrac{P(A \text{ and } B)}{P(A)} \)
Two hundred fifty people who recently purchased a car were questioned and the results are summarized in the following table from Section 5.3.
| Satisfied | Not Satisfied | Total | |
|---|---|---|---|
| New Car | 92 | 28 | 120 |
| Used Car | 83 | 47 | 130 |
| Total | 175 | 75 | 250 |
This is a conditional probability because we already know that the person bought a used car.What is the probability that a person is satisfied if it is known that the person bought a used car?
Solution
\[ \begin{align*} P(\text{satisfied | used car}) &= \dfrac{P(\text{satisfied and used})}{P(\text{used})} \\[4pt]&= \dfrac{\dfrac{83}{250}}{\dfrac{130}{250}} = \dfrac{83}{\cancel{250}} \cdot \dfrac{\cancel{250}}{130} = \dfrac{83}{130} \approx 0.638 \end{align*}\]
The probability that a person is satisfied if it is known that the person bought a used car is approximately 0.638 or 63.8%
A survey of 350 students at a university revealed the following data about class standing and place of residence.
Table \(\PageIndex{2}\): Housing by Class
|
Residence\Class |
Freshman |
Sophomore |
Junior |
Senior |
Row Totals |
|---|---|---|---|---|---|
|
Dormitory |
89 |
34 |
46 |
15 |
184 |
|
Apartment |
32 |
17 |
22 |
48 |
119 |
|
With Parents |
13 |
31 |
3 |
0 |
47 |
|
Column Totals |
134 |
82 |
71 |
63 |
350 |
What is the probability that a student is a sophomore if that student lives in an apartment?
Solution
This is a conditional probability because we are given that the student lives in an apartment.
\[ \begin{align*} P(\text{sophomore | apartment}) &= \dfrac{P(\text{sophomore and apartment})}{P(\text{apartment})} \\[4pt]&= \dfrac{\dfrac{17}{350}}{\dfrac{119}{350}} = \dfrac{17}{\cancel{350}} \cdot \dfrac{\cancel{350}}{119} = \dfrac{17}{119} \approx 0.143 \end{align*}\]
The probability that a student is a sophomore given that the student lives in an apartment is approximately 0.143 or 14.3%.