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.
X
0 1 2 3
Y 0 K 0.02 0.01 0.03 1
1 0.01 0.03 0.05 0.04 2
2 0.02 0.01 0.04 0.69
A restaurant needs a staff of 3 waiters and 2 chef
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?

Answer 1

Answer:

a) The value of k that mass this a valid pribability model = 0.04

b) The probability that at least one waiter and at least one chef show up on any given day = 0.86

c) The probability that more chefs show up than waiters on any given day = 0.04

d) The probability that more than three total staff (waiters and chefs) will show up on any given day = 0.77

Step-by-step explanation:

The table of probabilities represents the combined probabilities of whatever number of waiters and chef.

The joint probability model for the number of waiters (X) and chefs (Y), including the 'total' row amd column, showing the sum of the joint probabilities along a row or column

Note That, for this to be a valid probability model, all of the joint probabilities must sum up to give 1.00

Y/X | 0 | 1 | 2 | 3 | Total

0 | K | 0.02 | 0.01 | 0.03 | (k+0.07)

1 | 0.01 | 0.03 | 0.05 | 0.04 | 0.13

2 | 0.02 | 0.01 | 0.04 | 0.69 | 0.76

T | (k+0.03) | 0.07 | 0.10 | 0.76 | 1.00

Note; (Y/X) just shows that Y is for the vertical and X takes the horizontal. And T = total along the vertical.

a) Using either the total along the vertical or horizontal, it should end up at the same answer of 1.00 if this is truly a valid probability model.

(k+0.07) + 0.13 + 0.76 = 1.00

or

(k+0.03) + 0.07 + 0.10 + 0.76 = 1.00

Either ways,

k = 1.00 - 0.96 = 0.04

The table then becomes

Y/X | 0 | 1 | 2 | 3 | Total

0 | 0.04 | 0.02 | 0.01 | 0.03 | 0.11

1 | 0.01 | 0.03 | 0.05 | 0.04 | 0.13

2 | 0.02 | 0.01 | 0.04 | 0.69 | 0.76

T | 0.07 | 0.07 | 0.10 | 0.76 | 1.00

b) Probability that at least one waiter and at least one chef show up on any given day = P(X≥1 n Y≥1)

This probability = P(X=1 n Y=1) + P(X=1 n Y=2) + P(X=2 n Y=1) + P(X=2 n Y=2) + P(X=3 n Y=1) + P(X=3 n Y=2)

= 0.03 + 0.01 + 0.05 + 0.04 + 0.04 + 0.69

= 0.86

c) Probability that more chefs show up than waiters on any given day = P(X n Y) where (Y > X)

This probability = P(X=0 n Y=1) + P(X=0 n Y=2) + P(X=1 n Y=2)

= 0.01 + 0.02 + 0.01

= 0.04

d) Probability that more than three total staff (waiters and chefs) will show up on any given day = P(X n Y) where (X + Y) > 3

This probability = P(X=2 n Y=2) + P(X=3 n Y=1) + P(X=3 n Y=2)

= 0.04 + 0.04 + 0.69

= 0.77

Hope this Helps!!!