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Assoli18 [71]
3 years ago
11

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?
Mathematics
1 answer:
mojhsa [17]3 years ago
4 0

Answer:

Follows are the solution to this question:

Step-by-step 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)\times 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

In point c:\to P(X+Y

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

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