Answer:
Assumption
Let A be the event that the person stops at first signal.
Let B be the event that the person stops at second signal.
Given
Probabilities:
P(A)=0.40
P(B)=0.50
P(A∪B)=0.50
(a) Stops at both signals
The probability that person stops at both signals is intersection of events A and B
P(A∩B) = P(A)+P(B)−P(A∪B)
= 0.4+0.5−0.6=0.9−0.6
=0.3
(b) Stops at first but not at second
The given case is intersection of event A wth compliment of B
P(A∩¯B) = P(A)−P(A∩B)
=0.40−0.30
=0.10
(c) Stops at exactly one signal
The given case is sum of events that when a person stops either only first signal or at only second signal.
P = [P(A)−P(A∩B)]+[P(B)−P(A∩B)]
= [0.4−0.3]+[0.5−0.3]
= [0.1]+[0.2]
= 0.10+0.20
=0.30
