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 0.4, 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 0.65. 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 on signal?

a) The probability is 0.3

b) The probability is 0.1

c) The probability is 0.35

Step-by-step explanation:

Lets call S1 and S2 the events ' he must stop at signal 1 ' and 'he must stop at signal 2' respectively.

a) We know that

0.65 = P(S1 ∪ S2) = P(S1) + P(S2) - P(S1 ∩ S2) = 0.4+0.55-P(S1 ∩ S2) = 0.95 - P(S1 ∩ S2)

Hence P(S1 ∩ S2) = 0.95-0.65 = 0.3

It stops at both signals with probability 0.3

b) Note that, due to the theorem of total probability we have

0.4 = P(S1) = P(S1 ∩ S2) + P(S1 ∩ S2^c) = 0.3 + P(S1∩S2^c)

Where S2^c is the complementary event of S2. Therefore

P(S1∩S2^c) = 0.4-0.3 = 0.1

The probability to stop at the first signal but not at the second one is 0.1

c) The probability of stopping at exactly one signal is equal at the sum of the probabilities of stopping only at the first signal and the probability of stopping only at the second one.

That is P(S1 ∩ S2^c) + P(S1^c ∩ S2) = 0.1 + P(S1^c ∩ S2)

The same way as before:

0.55 = P(S2) = P(S1 ∩ S2) + P(S1^c ∩ S2) = 0.3 + P(S1^c ∩ S2)

Therefore

P(S1^c ∩ S2) = 0.55-0.3 = 0.25

And as a result, the probability of stopping at exactly one signal is 0.25 + 0.1 = 0.35.

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step-by-step explanation: