Section 8-3 Conditional Probability, Intersection, and Independence
Example 1. A pair of dice are rolled.
a. What is the probability that the dot sum is a prime number given that it
is greater than 7?
PA | B 
2
PA  B
P(prime   7)
2
 P(prime |  7) 
 36 
15
PB
P( 7)
15
36
b. What is the probability that the dot sum is greater than 7 given that it is a prime
number?
Prime
P( 7 | prime) 
2
P( 7  prime)
2
 36 
15
P(prime)
15
36
>7
Yes (2/15)
Prob
2/36
Yes (15/36)
No (13/15) 13/36
Yes (13/21) 13/36
No (21/36)
No (8/21)
8/36
Example 2. The spinner shown below is spun.
Even/Odd
Prime/Not
Prime (2/5)
Prob
5/8 * 2/5 = 2/8
Even (5/8)
Not Prime (3/5) 5/8 * 3/5 = 3/8
Prime (1/3)
3/8 * 1/3 = 1/8
Odd (3/8)
a. What is the probability that the spinner result will be
a prime number given that it is an even number?
Not Prime (2/3) 3/8 * 2/3 = 2/8
2
P(prime  even)
2
P(prime | even) 
 8 
5
P(even)
5
8
b. What is the probability that the spinner result will be an even number given that it is a
prime number?
P(even | prime) 
2
P(even  prime)
2
 8 
3
P(prime)
3
8
Example 3. An automobile manufacturer produces 37% of its cars at plant A. If 5% of the cars
manufactured at plant A have defective emission control devices, what is the probability that one of this
manufacturer’s cars was manufactured at plant A and has a defective emission control device?
P(Plant A  Defective)  P(Plant A)P(Defect ive | Plant A)  0.37  0.05  0.0185
Plant
Defective Prob
Yes (0.05) 0.0185
A (0.37)
No (0.95) 0.3515
Example 4. To transfer into a particular technical department, a company requires an
employee to pass a screening test. A maximum of 3 attempts are allowed and an
employee must wait at least 6 months before repeating the test. From past records it is
found that 40% pass on the first trial. Of those who fail the first test and repeat the test,
60% pass. Of those who take the test the third time, 20% pass. For an employee wishing
to transfer:
Try 1
Pass (0.4)
Try 2
Pass (0.6)
Fail (0.6)
Try 3
Prob
0.400
0.360
Pass (0.2) 0.048
Fail (0.4)
Fail (0.8) 0.192
a. What is the probability of passing the test on the first or second try?
P(Try 1  Try 2)  P(Try 1)  P(Try 2)  0.400  0.360
b. What is the probability of failing on the first two tries and passing on the third?
P((Fail 1  Fail 2)  Pass 3)  P(Fail 1  Fail 2)P(Pass 3 | Fail 1  Fail2)
 P(Fail1  Fail 2)  0.2
 P(Fail 1)P(Fail 2 | Fail 1)  0.2
 0.6  0.4  0.2
 0.048
c. What is the probability of failing on all 3 attempts?
P((Fail 1  Fail 2)  Fail 3)  P(Fail 1  Fail 2)P(Fail 3 | Fail 1  Fail2)
 P(Fail1  Fail 2)  0.8
 P(Fail 1)P(Fail 2 | Fail 1)  0.8
 0.6  0.4  0.8
 0.192
Example 5. In a study to determine frequency and dependency of color blindness relative
to females and males, 1,000 were chosen at random and the following results recorded:
Female Male
F
M Total
Color-Blind C
2
24
26
Normal C’
518 456
974
Total
520 480 1000
Gender
Color-Blind
Yes (24/480)
Prob
0.024
No (456/480)
0.456
Yes (2/520)
0.002
Color-Blind
Gender
Male (24/26)
Prob
0.024
Female (2/26)
0.002
Male (456/974)
0.456
Yes (26/1000)
Male (480/1000)
No (974/1000)
Female (520/1000)
No (518/520)
0.518
Female (518/974) 0.518
a. What is the probability that a person is a woman, given that the person is color blind?
P(Female | Color Blind) 
2
P(Female  Color - Blind)
2
 1000 
26
P(Color - Blind)
26
1000
b. What is the probability that a person is color-blind, given that the person is a male?
P(Color - Blind | Male) 
24
P(Color - Blind  Male)
1000  24

480
P(Male)
480
1000
c. Are the events color-blindness and male independent?
In general, A and B are independent if P(A) = P(A|B) or, equivalently, P(B) = P(B|A).
26
1000
No, because P(Color-Blind) ≠ P(Color-Blind|Male)
P(Color - Blind) 
d. Are the events color-blindness and female independent?
In general, A and B are independent if P(A) = P(A|B) or, equivalently, P(B) = P(A|B).
520
P(Female) 
1000
No, because P(Female) ≠ P(Female|Color-Blind)