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Chapter 2: Q22E (page 65)

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 .4, the analogous probability for the second signal is .5, and the probability that he must stop at at least one of the two signals is .7. 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?

Short Answer

Expert verified

a. The probability that he must stop at both signals is 0.2.

b. The probability that he stops at the first signal but not at the second one is 0.2.

c. The probability that he stops at exactly one signal is 0.5.

Step by step solution

01

Given information

The probability that a motorist stopped at the first signal is 0.4.

The probability that a motorist stops at the second signal is 0.5.

The probability that must stop at at least one of the two signals is 0.7.

02

Compute the probability

Let A be the event that a motorist stopped at the first signal.

Let B be the event that a motorist stops at the second signal.

From the provided information, the probabilities are given as,

\(\begin{aligned}P\left( A \right) = 0.4\\P\left( B \right) = 0.5\\P\left( {A \cup B} \right) = 0.7\end{aligned}\)

a.

Using the formula for the probability of the union of the two events,

\(P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\)

The probability that he must stop at both signals is computed as,

\(\begin{aligned}P\left( {A \cap B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)\\ &= 0.4 + 0.5 - 0.7\\ &= 0.2\end{aligned}\)

Therefore, the probability that he must stop at both signals is 0.2.

b.

The probability that he stops at the first signal but not at the second one is computed as,

\(\begin{aligned}P\left( {A \cap B'} \right) &= P\left( A \right) - P\left( {A \cap B} \right)\\ &= 0.4 - 0.2\\ &= 0.2\end{aligned}\)

Therefore, the probability that he stops at the first signal but not at the second one is 0.2.

c.

The probability that he stops at the second signal but not at the first one is computed as,

\(\begin{aligned}P\left( {A' \cap B} \right) &= P\left( B \right) - P\left( {A \cap B} \right)\\ &= 0.5 - 0.2\\ &= 0.3\end{aligned}\)

Therefore, the probability that he stops at the second signal but not at the first one is 0.3.

The sum of the above computed probability and the probability computed in (b) part is the probability that he stops at exactly one signal.

The probability that he stops at exactly one signal is computed as,

\(\begin{aligned}P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) &= 0.2 + 0.3\\ &= 0.5\end{aligned}\)

Therefore, the probability that he stops at exactly one signal is 0.5.

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