15(b – 1) + 2(3b + 7) = 6

a) P(X∩Y) = 0.2

b) = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability that he must stop at the first signal but not at the second one can be calculated as:

= P(X) - P(X∩Y)

= 0.36 - 0.2 = 0.16

At the same way, the probability that he must stop at the second signal but not at the first one can be calculated as:

= P(Y) - P(X∩Y)

= 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is: