A large but sparsely populated county has two small hospitals, one at the south

Question

A large but sparsely populated county has two small hospitals, one at the south end of the county and the other at the north end. The south hospital’s emergency room has 4 beds, whereas the north hospital’s emergency room has only 3 beds. Let X denote the number of south beds occupied at a particular time on a given day, and let Y denote the number of north beds occupied at the same time on the same day. Suppose that these two rvs are independent, that the pmf of X puts probability masses .1, .2, .3, .3, and .2 on the x values on the x values 0, 1, 2, 3, and 4, respectively, and that the pmf of Y distributes probabilities .1, .3, .4, and .2 on the y values 0, 1, 2, and 3, respectively.

a) Display the joint pmf of X and Y in a joint probability table.

b) Compute P(X <= 1 and Y <= 1) by adding proabilities from the joint pmf, and verify that this equals the product of P(X <= 1 ) and P(Y <= 1 ).

c) Express the event that the total number of beds occupied at the two hospitals combined is at most 1 in terms of X and Y, and then calculate this probability.

d) What is the probability that at least one of the two hospitals has no beds occupied?

Explanation / Answer

a) Since X and Y are indpendent, P(X=x, Y=y) = P(X=x)*P(Y=y)

The jpmf of (X ,Y) is

b)

P(X <= 1 and Y <= 1) = 0.01+ 0.03+ 0.02+ 0.06 = 0.12

P(X<=1) = P(X=1) + P(X=0) = 0.1+0.2 = 0.3

P(Y<=1) =P(Y=1) +P(Y=0) = 0.1+ 0.3 = 0.4

Now, P(X<=1) P(Y<=1) = 0.3*0.4 = 0.12 = P(X <= 1 and Y <= 1) since x and y are indepdent

c) P(X+Y <=1) = P(X=0,Y=0) + P(X=1,Y=0) + P(X=0,y=1)

= 0.01+0.03 + 0.02 = 0.06

d) The probability that at least one of the two hospitals has no beds occupied is

P(X=0,Y=0) + P(X=1,Y=0) + P(X=0,y=1) = 0.01+0.03 + 0.02 = 0.06

X Y 0 1 2 3 P(X=x) 0 0.01 0.03 0.04 0.02 0.1 1 0.02 0.06 0.08 0.04 0.2 2 0.03 0.09 0.12 0.06 0.3 3 0.03 0.09 0.12 0.06 0.3 4 0.01 0.03 0.04 0.02 0.1 P(Y=y) 0.1 0.3 0.4 0.2 1

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