A restaurant needs a staff of 3 waiters and 2 chefs to be properly staffed. The

Question

A restaurant needs a staff of 3 waiters and 2 chefs to be properly staffed. The joint probability model for the number of waiters (X) and chefs (Y) that show up on any given day is given below. a) What must the value of k be for this to be a valid probability model? b) What is the probability that at least one waiter and at least one chef show up on any given day? c) What is the probability that more chefs show up than waiters on any given day? d) What is the probability that more than three total staff (waiters and chefs) will show up on any given day? e) What is the expected total number of staff (waiters and chefs) that will show up on any given day? f) What is the probability that three waiters will show up on any given day? g) What is the probability that two chefs will show up on any given day? X and Y are independent. False True

Explanation / Answer

Let's assume, probability that waiter shows up = P(W)

probability that chef shows up = P(C)

Solution A:

Sum of the probabilities for all cells must add to 1. Hence, value of k = 1-0.93 = 0.07

Solution B:

Probability that at least 1 waiter and 1 chef show up =

P(1W and 1C) + P(1W and 2C) + P(2W and 1C) + P(2W and 2C) + P(3W and 1 C) + P(3W and 2C)

From the above table, substituting the values in above equation:

= 0.03 + 0.02 + 0.04 + 0.06 + 0.05 + 0.61 = 0.81

Hence, Probability that at least 1 waiter and 1 chef show up = 0.81

Solution C:

Probability that more chefs show up than waiter =

P(1C and 0W) +P(2C and 0W) + P(2C and 1W)

From the above table, substituting the values in above equation:

= 0.02+ 0.02 + 0.02 = 0.06

Hence, Probability that more chefs show up than waiters = 0.06

Solution D:

Probability that more than 3 total staff will show up =

P(2W and 2C) +P(3W and 1C) + P(3W and 2C)

From the above table, substituting the values in above equation:

= 0.06+ 0.05 + 0.61= 0.72

Hence, Probability that more than 3 staff will show up = 0.72

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