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. What must the value of k be for this to be a valid probability model? What is the probability that at least one waiter and at least one chef show up on any given day? What is the probability that more chefs show up than waiters on any given day? What is the probability that more than three total staff (waiters and chefs) will show up on any given day? What is the expected total number of staff (waiters and chefs) that will show up on any given day? What is the probability that three waiters will show up on any given day? What is the probability that two chefs will show up on any given day? X and Y are independent. False True

Explanation / Answer

(a)

To find the numerical value of k we will use that the fact that sum of probabilities of all events is always 1. Therefore

k+0.02+0.02+0.02+0.01+0.04+0.05+0.04+0.01+0.03+0.05+0.67 = 1

thus k = 0.04

(b)

P(X>=1,Y>=1 ) = 0.04+0.02+0.01+0.04 = 0.11

(c)

P(Y>X) = 0.01+0.01+0.03 =0.05

(d)

P(X+Y>3) = 0.04+0.05+0.67 = 0.76

(e)

E(X+Y) = 0*04+1*0.02+2*0.02+3*0.02+1*0.01+2*0.04+3*0.05+4*0.04+2*0.01+3*0.03+4*0.05+5*0.67 = 4.18

(f)

P(X = 3) = 0.02+0.04+0.67 = 0.73

(g)

P(Y = 2) = 0.01+0.03+0.05+0.67 = 0.76

We have,

P(X=3 and Y=2) = 0.67

Also, P(X=3)*P(Y=2) = 0.73*0.76 = 0.55

Since the above multiplication is not equal to the probability of the events happening together, X and Y are not independent,

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