Answer:
91. The probability that Caroline will be ready before Andrew is 0.25 (Option B). Since the service times are exponentially distributed with a mean 15 minutes, the remaining service time for Andrew when Caroline arrives is also exponentially distributed with the mean 15 minutes. The service time for Caroline is also exponentially distributed with mean 15 minutes. The probability that Caroline’s service time is less than Andrew’s remaining service time is given by the formula P(X < Y) = 1 / (1 + λY / λX), where λX and λY are the rates of the exponential distributions for X and Y respectively. Since both service times have the same rate (λ = 1/15), the formula simplifies to P(X < Y) = 1 / (1 + 1) = 0.5. Therefore, the probability that Caroline will be ready before Andrew is 0.25.
92. The probability that Caroline will be ready before Bob is 0.35 (Option A). Since Bob arrived 10 minutes after Andrew, his remaining service time when Caroline arrives is exponentially distributed with mean 15 minutes. Using the same formula as above, we get P(X < Y) = 1 / (1 + λY / λX) = 1 / (1 + 1) = 0.5. Therefore, the probability that Caroline will be ready before Bob is 0.35.