Question

The route used by a certain motorist in commuting to work contains two intersections with traffic signal lights. The probability that she must stop at the first signal and second signal are 0.40 and 0.50, respectively. The probability that she must stop at either signal is 0.60. What is the probability that she must stop at the first signal but not the second signal? Let: F = must stop at first signal F’ = do not have to stop at first signal S = must stop at second signal S’ = do not have to stop at second signal

Answer 1

Answer: 0.1

Step-by-step explanation:

Let F = must stop at first signal .

F’ = do not have to stop at first signal .

S = must stop at second signal .

S’ = do not have to stop at second signal.

As given , we have

P(F)=0.40

P(S)=0.50

P(F∪S)=0.60

Use formula , $P(F\cap S)=P(F)+P(S)-P(F\cup S)$

i.e. $P(F\cap S)=0.40+0.50-0.60=0.30$

The probability that she must stop at the first signal but not the second signal :

$P(F\cap S')=P(F)-P(F\cap S)=0.40-0.30=0.10$

Hence, the probability that she must stop at the first signal but not the second signal= 0.1