Question

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 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) 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? e. What is the probability that the south end has two or more beds occupied given that the north end has 2 beds occupied?

Answer 1

Let's address each part of this problem step by step:

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

To create a joint probability table, we list all possible combinations of X and Y and calculate their probabilities. Given that X and Y are independent, we can multiply the probabilities of X and Y together to find the joint probabilities:

X And P(X, Y)

1 0 0.01

1 1 0.03

1 2 0.04

1 3 0.02

2 0 0.02

2 1 0.06

2 2 0.08

2 3 0.04

3 0 0.03

3 1 0.09

3 2 0.12

3 3 0.06

4 0 0.02

4 1 0.06

4 2 0.08

4 3 0.04

b. Compute P(X≤1 and Y≤1):

To find this probability, we sum the joint probabilities where X is less than or equal to 1 and Y is less than or equal to 1:

P(X≤1 and Y≤1) = P(X=1, Y=0) + P(X=1, Y=1) + P(X=0, Y=1)

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

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:

The event that the total number of beds occupied is at most 1 can be expressed as (X + Y ≤ 1). To calculate this probability, we sum the joint probabilities where X + Y is less than or equal to 1:

P(X + Y ≤ 1) = P(X=1, Y=0) + P(X=0, Y=1) + P(X=0, Y=0)

P(X + Y ≤ 1) = 0.01 + 0.02 + 0.1 = 0.13

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

To find the probability that at least one of the two hospitals has no beds occupied, we can calculate the complementary event where both hospitals have beds occupied:

P(at least one hospital with no beds occupied) = 1 - P(both hospitals have beds occupied)

P(both hospitals have beds occupied) = P(X > 0 and Y > 0)

P(both hospitals have beds occupied) = 1 - P(X=0, Y=0) = 1 - 0.1 = 0.9

So, the probability that at least one of the two hospitals has no beds occupied is 0.9.

and. What is the probability that the south end has two or more beds occupied given that the north end has 2 beds occupied?

To find this conditional probability, we need to calculate the probability that X is greater than or equal to 2 given that Y is equal to 2:

P(X ≥ 2 | Y = 2) = P(X = 2 | Y = 2) + P(X = 3 | Y = 2) + P(X = 4 | Y = 2)

P(X = 2 | Y = 2) = 0.08

P(X = 3 | Y = 2) = 0.12

P(X = 4 | Y = 2) = 0.08

P(X ≥ 2 | Y = 2) = 0.08 + 0.12 + 0.08 = 0.28

So, the probability that the south end has two or more beds occupied given that the north end has 2 beds occupied is 0.28.

Explanation: