Question

The hotel manager loses the key labels for three rooms (1, 2, 3). That is, he does not know which doors the keys can open. He randomly distributes the three keys to three rooms' guests. Define random variable X: the number of guest who receives the correct key. (1)(4pts) Determine the pmf of X. (2)(3pts) Find the probability that at most two guests receive the correct keys. (2)(3pts) Calculate the expected va

lue and variance of random variable X.

Answer 1

Let us see at all the combinations of keys with the rooms and the compute the value of X in each case as:

Room-1	Room-2	Room-3	X
1	2	3	3
1	3	2	1
2	1	3	1
2	3	1	0
3	1	2	0
3	2	1	1

a) Using the above table, the PMF for X here is obtained as:
P(X = 0) = 2/6 = 1/3
P(X = 1) = 3/6 = 1/2
P(X = 3) = 1/6

This is the required PMF for X here.

b) Probability that at most 2 guests receives the correct key is computed here as:
P(X <= 2) = 1- P(X = 3) = 1 - (1/6) = 5/6

Therefore 5/6 is the required probability here.

c) The expected value of X here is obtained as:
E(X) = 0*(1/3) + 1*(1/2) + 3*(1/6) = 1
Therefore 1 is the expected value of X here.

The second moment of X here is obtained as:
E(X2) = 02*(1/3) + 12*(1/2) + 32*(1/6) = (1/2) + (9/6) = 2

Therefore, Var(X) = E(X2) - [E(X)]2 = 2 - 12 = 1

Therefore 1 is the required Variance of X here.