anyone good at binomial distibution?

Question

1/5 of the female freshmen at a private college are out of state students. if the students are assigned at random to the dormitories, 3 to a room, how do i find the probability that, in one room, at most 2 of the 3 roomates are out of state students?

Answer 1

I would solve this by figuring out the probability that exactly 3 of the 3 roommates are out of state, then subtracting from 1.

P(3 of the roommates are out of state) = 1/5^3 = 1/125

So the probability that at most 2 are out of state = 124/125

The long way would be to figure out the individual probabilities for 0 roommates, 1 roommate and 2 roommates.

Let p be the probability that the female student is from out of state (1/5).
Let 1-p be the probability that the female student is in state (4/5).

The probability that exactly k of the n women in a room are from out of state is:
P( out of state = k ) = C(n,k) * p^k * (1-p)^(n-k)

Let's figure this out for each of the values k = 0, 1, 2:

P( 0 women from out of state )
= C(3,0) * (1/5)^0 * (4/5)^3
= 1 * 1 * (4/5)^3
= 64/125

P( 1 woman from out of state )
= C(3,1) * (1/5)^1 * (4/5)^2
= 3 * 1/5 * (4/5)^2
= 3 * 16 / 125
= 48 / 125

P( 2 women from out of state )
= C(3,2) * (1/5)^2 * (4/5)^1
= 3 * (1/5)^2 * 4/5
= 3 * 4/ 125
= 12 / 125

Adding them up:
64/125 + 48/125 + 12/125
= (64 + 48 + 12) / 125
= 124 / 125

Whew! We get the same answer either way! Again, I vote for the first method. :-)

Answer 2

This binomial has n = 3 and p = 0.20
In general, P(x) = n!/(x!(n-x)!) p^x(1-p)^(n-x)
P(x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2)
= 3!/(0!(3!)0.20^0(0.80)^3 + 3!/(1!(2!)0.20^1(0.80)^2 + 3!/(2!(1!)0.20^2(0.80)^1
= 0.512 + 0.384 + 0.0960 = 0.992 (answer)