A large but sparsely populated county has two small hospitals, one at the south end of the county and one at the north end. The south hospital emergency room has four beds, whereas the north hospital emergency room has only 3 beds. Let X denote the number of south beds occupied at a particular time on a given day. Let Y denote the number of North beds occupied at the same time on the same day. Suppose that these two random variables are independent; that the pmf of X puts probability masses .1, .2, .3, .2 and .2 on x values of 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 probabilities 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?

Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by

f(x,y) = P((X=x,Y=y) ,

and given that the random variables are independent the joint pmf isbgiven by:

f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)

(b) the required probability is given by considering X=0,1 and Y =0,1

f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }

= [0.1 +0.2 ] × [0.1 + 0.3]

= 0.3 × 0.4

= 0.12

Now we find

fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1

fX(x) = sum{[f(x,y)]}= 0.1+0.3

= 0.4

fY(y) = sum{[f(x.y)]} = 0.1 + 0.2

= 0.3

Since fY(y) × fX(x) = 0.4 × 0.3

= 0.12

Hence f(x,y) = fX(x) .fY(y), the events are independent

(c) the required event is give by: [P(X<=1, PY<=1)]

= P(X<=1) . P(Y<=1)

= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1

= 0.4 × 0.3

= 0.12

(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A

A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]

A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]

A = [0.09] + [ 0.09]

A = 0.18

Pls note the sum of the vertical column X=2 is 0.3