A commercial building has four entrances numbered IIIIIIIV T

A commercial building has four entrances, numbered I,II,III,IV. Three people enter the building at 9:00 a.m. Let X denote the number who selected entrance I. Assuming that the people choose entrances independently and at random, find the probability distribution for X. Were any additional assumptions necessary for your answer?

Solution

There are 4^3 = 64 ways for the three people to choose entrances.

If X = 0, there are 3^3 = 27 ways for them to choose entrances.

If X = 1, there are 3 persons who can do this. There are 3^2 = 9 ways for the other two to choose entrances. Hence, there are 3*9 = 27 ways for X = 1.

If X = 2, there are 3C2 = 3 pairs who can do this. There are 3 ways for the last person to choose entrances. Hence, there a re 3*3 = 9 ways for X= 2.

There is only 1 ways so that X = 3.

Thus, the probability distirbution is

P(x) = 27/64, x = 0,

27/64, x = 1
9/64, x = 2
1/64, x = 3 [ANSWER]
0, otherwise

The assumption is that the three people choose their entrances randomly.