Statistics - Practice Problems
Mutually exclusive or disjoint events, dependent and independent events (Nonempty events A and B)
Mutually exclusive or disjoint events:
S
P( A and B )  0
A
A
B
P( A or B )  P( A )  P(B )
P(A) + P(B) < 1
S is a sample space.
Events A and B are mutually exclusive or
disjoint events.
S
A
B
P(A|B)=0
P(B|A)=0
Not Mutually exclusive or disjoint
events, the events overlap.
P( A and B )  0
A and B
S is a sample space.
Events A and B are mutually exclusive or
disjoint events.
P( A or B )  P( A )  P(B )  P( A and B )
Note: P(A) + P(B) does not have to be <1
For independent events:
P(A and B) = P(A)P(B)
For non-independent events
P(A and B) = P(A)P(B|A)
P( A and B )
And P( A | B ) 
P( B )
Problem 1.
1. P(A) = 0.7, P(B) = 0.4, P(A and B) = 0.3
a) Are A and B mutually exclusive?
No, because P(A and B)  0
b) Are A and B independent events?
No, because P(A and B)  P(A)P(B), that is, 0.3  (0.7)(0.4)
c) Find P(A or B)
P(A or B) = P(A) + P(B) – P(A and B) = 0.7 + 0.4 – 0.3 = 0.8
d) Find P(A|B)
P( A and B) 0.3
P( A | B ) 

 0.75
P( B )
0.4
Problem 2.
1. P(A) = 0.7, P(B) = 0.4, P(A and B) = 0.28
a) Are A and B mutually exclusive?
No, because P(A and B)  0
b) Are A and B independent events?
Yes, because P(A and B) =P(A)P(B), that is, 0.28 (0.7)(0.4)
c) Find P(A or B)
P(A or B) = P(A) + P(B) – P(A and B) = 0.7 + 0.4 – 0.28 = 0.82
d) Find P(A|B)
P( A and B ) 0.28
P( A | B ) 

 0.7  P( A )
P( B )
0.4
Problem 3.
Only two students in the class are independently trying to solve a very difficult probability
problem. Student A has a probability of 0.9 of solving the problem while student B has a
probability of 0.6 of solving the problem.
a) What is the probability that both of them will solve the problem?
Solution:
Because they are working independently, P(A and B) = P(A)P(B) = (0.9)(0.6) = 0.54
b) What is the probability that both of them fail to solve the problem?
Solution:
P(A fails to solve the problem and B fails to solve the problem)
P( A  and B )  P( A  )( P(B ) =(1 - 0.9)(1 - 0.6) = (0.1)(0.4) = 0.04
d) What is the probability that the problem will be solved?
Solution:
The problem is solve if either one or both student solve the problem.
P(A or B) = P(A) + P(B) - P (A and B) = 0.9 + 0.6 – 054 = 0.96
d) What is the probability that only A solves the problem?
Solution:
P(only A) = P(A and B’) = P(A)P(B’) = (0.9)(0.4) = 0.36
e) What is the probability that only one of the students solves the problem?
Solution:
P(only one) = P(only A or only B) = P(A and B’) + P(A’ and B)
= (0.9)(0.4) + (0.1)(0.5)
= 0.36 + 0.05
= 0.41