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 four beds, whereas the north hospital's emergency room has only three 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 rv's are independent; that the pmf of X puts probability masses 0.2, 0.2, 0.3, 0.2, and 0.1 on the x values 0, 1, 2, 3, and 4, respectively; and that the pmf of Y distributes probabilities 0.4, 0.3, 0.2, and 0.1 on the y values 0, 1, 2, and 3, respectively.

a. Display the joint pmf of X and Yin a joint probabil- ity table.
b. Compute POX 1 and YS 1) by adding probabilities from the joint pmf, and verify that this equals the product of PX S 1) and POY S 1).
c. Express the event that the total number of beds occu- pied at the two hospitals combined is at most 1 in terms of X and and then calculate this probability.
d. What is the probability that at least one of the two hospitals has no beds occupied?

Answer 1

Follows are the solution to this question:

Explanation:

In point a:

Although both variables are non - stationary, for both the respective pmf, multiply all pmf principles. For example:

`\to P(x=0,y=0)=P(x=0)* P(y=0)`

`X`

`0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ pmf Y`

`0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.01 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.02 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.03\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.02\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.02\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1`

`Y \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ 0.03 \ \ \ \ \ \ \ \ \ \ \ 0.06 \ \ \ \ \ \ \ \ \ \ \ 0.09 \ \ \ \ \ \ \ \ \ \ \ 0.06 \ \ \ \ \ \ \ \ \ \ \ 0.06 \ \ \ \ \ \ \ \ \ \ \ 0.3`

`2 \ \ \ \ \ \ \ \ \ 0.04\ \ \ \ \ \ \ \ \ 0.08\ \ \ \ \ \ \ \ \ 0.12 0.08\ \ \ \ \ \ \ \ \ 0.08\ \ \ \ \ \ \ \ \ 0.4\\\\ 3\ \ \ \ \ \ \ \ \ 0.02\ \ \ \ \ \ \ \ \ 0.04\ \ \ \ \ \ \ \ \ 0.06\ \ \ \ \ \ \ \ \ 0.04\ \ \ \ \ \ \ \ \ 0.04\ \ \ \ \ \ \ \ \ 0.2\\\\pmf X \ \ \ \ \ \ \ \ \ 0.1\ \ \ \ \ \ \ \ \ 0.2\ \ \ \ \ \ \ \ \ 0.3\ \ \ \ \ \ \ \ \ 0.2\ \ \ \ \ \ \ \ \ 0.2``

In point b:

`\to P(x<=1,y<=1)=0.01+0.02+0.03+0.06=0.12\\\\\to P(x<=1)=0.1+0.2=0.3 \\\\\to P(y<=1)=0.1+0.3=0.4\\\\\to P(x<=1 \ and\ y<=1)=P(x<=1) * P(y<=1)=0.3 * 0.4=0.12 \\\\`

In point c:
`\to P(X+Y<=1)=P(X=0,Y=1)+(X=1,Y=0)+P(X=0,Y=0)=0.02+0.03+0.01=0.06`

In point d:

`\to P(X=0,Y=1)+P(X=0,Y=2)+P(X=0,Y=3)+P(X=0,Y=0)+P(X=1,Y=0)+P(X=2,Y=0)+P(X=3,Y=0)+P(X=4,Y=0) \\\\ \to 0.03+0.04+0.02+0.01+0.02+0.03+0.02+0.02=0.19`