Answer to Question #152651 in Statistics and Probability for ali

Question

A traffic survey on a busy highway reveals that one of every four vehicles is a truck. This survey also established that one-eighth of all automobiles are unsafe to drive and one-twentieth of all trucks are unsafe to drive.
a. What i s the probability that the next vehicle to pass a given point i s an unsafe truck?
b. What is the probability that the next vehicle will be a truck, given that it is unsafe?
c. What is the probability that the next vehicle that passes a given point will be a truck, given that the previous vehicle was an automobile?

Answer 1

P(T) = probability of truck

P(U) = probability of unsafe vehicle

$P(T) = \frac{1}{4} \\ P(R) = \frac{3}{4} \\ P(U) = \frac{1}{8} \\ P(U/T) = \frac{1}{20} \\ P(U) = P(T)P(U/T) + P(R)P(U/R) \\ \frac{1}{8} = \frac{1}{4} \times \frac{1}{20} + \frac{3}{4}P(U/R) \\ P(U/R) = \frac{3}{20}$

a.

$P(U \cap T) = P(T)P(U/T) \\ = \frac{1}{4} \times \frac{1}{20} \\ = \frac{1}{80} \\ = 0.0125$

b.

$P(T/U) = \frac{P(U \cap T)}{P(U)} \\ = \frac{1}{80} \times \frac{8}{1} \\ = \frac{1}{10} \\ = 0.1$

c.

$P(T/A) = \frac{P(T \cap A)}{P(A)} \\ = \frac{P(T)}{P(A)} \\ = \frac{\frac{1}{4}}{1} \\ = \frac{1}{4} \\ = 0.25$