Question

For 91-92; A dental surgery has two operation rooms. The service times are assumed to be independent, exponentially distributed with mean 15 minutes. Andrew arrives when both operation rooms are empty. Bob arrives 10 minutes later while Andrew is still under medical treatment. Another 20 minutes later Caroline arrives and both Andrew and Bob are still under treatment. No other patient arrives during this 30-minute interval. 91. What is the probability that Caroline will be ready before Andrew? A. 0.35 B. 0.25 C. 0.52 D. None of these 92. What is the probability that Caroline will be ready before Bob? A. 0.35 B. 0.25 C. 0.52

Answer 1

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.