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?

Step-by-step explanation:

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

it is part of the rising action that reveals public sentiment toward hester.

explanation:

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.