Question

2. 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 0.36, the analogous probability for the second signal is 0.54, and the probability that he must stop at at least one of the two signals is 0.65. What is the probability that he must stop at exactly one signal?

Answer 1

0.30

Explanation:

Probability of stopping at first signal = 0.36 ;

P(stop 1) = P(x) = 0.36

Probability of stopping at second signal = 0.54;

P(stop 2) = P(y) = 0.54

Probability of stopping at atleast one of the two signals:

P(x U y) = 0.6

Stopping at both signals :

P(xny) = p(x) + p(y) - p(xUy)

P(xny) = 0.36 + 0.54 - 0.6

P(xny) = 0.3

Stopping at x but not y

P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06

Stopping at y but not x

P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24

Probability of stopping at exactly 1 signal :

P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30