What is the probability that he will need to stop: (a) (b) (c) (d) <-- Listed in description

Question

The path used by a certain Canada Post mailman to deliver mail to the store includes two intersections with traffic signals. Probability of the event that he will have to stop at the first intersection is 0.35; and the probability that he will need to stop at the second one is 0.65. Furthermore, probability of making a stop at at least one of the two intersections is 0.7. What is the probability that he will need to stop:

(a) at both intersections?

(b) at the second intersection but not at the first one?

(c) at exactly one intersection?

(d) at the second intersection, given that he stopped at the first one? Is it different from 0.65 – why or why not?

Answer 1

Let the first intersection be F and the second be S

a) Often, we see the formula P(F ∪ S) = P(F) + P(S) - P(F ∩ S) or P(F) + P(S) -P(FS) (the first inclusion exclusion rule) .

but we can rewrite this

P(FS) = P(F) + P(S) - P(F ∪ S) = .35+ .65 - .7 = .3

b) P(F'S) = P(F ∪ S) - P(F) = .7 - .35 = .35

c)P(exactly one intersection) = P(F ∪ S) - P(FS) = .7 - .3 = .4 (you could either consider all of the components of F ∪ S or consider that the P(exactly one) = P(at least one) - P(exactly two), as if you have at least one, you must have one or two.

d)P(S|F) = P(FS)/P(F) = .3/.35 = 6/7