Question

The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.35, the analogous probability for the second signal is 0.55, and the probability that he must stop at at least one of the two signals is 0.75.
What is the probability that he must stop:

A. at both signals?
B. at the first signal but not at the second one?
C. at exactly one signal?

Answer 1

Answer:

a) 0.15

b) 0.2

c) 0.6

Step-by-step explanation:

We are given the following in the question:

A: Stopping at first signal

B: Stopping at second signal

P(A) = 0.35

P(B) = 0.55

Probability that he must stop at at least one of the two signals is 0.75

$P(A\cup B) = 0.75$

a) P(at both signals)

$P(A\cup B) = P(A) + P(B) - P(A\cap B)\\0.75 = 0.35 + 0.55 - P(A\cap B)\\P(A\cap B) = 0.35 + 0.55 - 0.75 = 0.15$

0.15 is the probability that motorist stops at both signals.

b) P(at the first signal but not at the second one)

$P(A\cap B') = P(A) - P(A\cap B)\\P(A\cap B') = 0.35 - 0.15 = 0.2$

0.2 is the probability that motorist stops at the first signal but not at the second one.

c) P(at exactly one signal)

$P(A\cap B') + P(A\cap 'B) = P(A\cup B) - P(A\cap B) \\P(A\cap B') + P(A\cap 'B) = 0.75 - 0.15 = 0.6$

0.6 is the probability that the motorist stops at exactly one signal.