Answer:
a) P(A∩B) = 0.29
b) P1 = 0.1
c) P = 0.35
Step-by-step explanation:
Let's call A the event that the motorist stop at the first signal, and B the event that the motorist stop at the second signal.
From the question we know:
P(A) = 0.39
P(B) = 0.54
P(A∪B) = 0.64
Where P(A∪B) is the probability that he stop in the first, the second or both signals. Additionally, P(A∪B) can be calculated as:
P(A∪B) = P(A) + P(B) - P(A∩B)
Where P(A∩B) is the probability that he stops at both signals.
So, replacing the values and solving for P(A∩B), we get:
0.64 = 0.39 + 0.54 - P(A∩B)
P(A∩B) = 0.29
Then, the probability P1 that he just stop at the first signal can be calculated as:
P1 = P(A) - P(A∩B) = 0.39 - 0.29 = 0.1
At the same way, the probability P2 that he just stop at the second signal can be calculated as:
P2 = P(B) - P(A∩B) = 0.54 - 0.29 = 0.25
Finally, the probability P that he stops at exactly one signal is:
P = P1 + P2 = 0.1 + 0.25 = 0.35
