Binomial Distribution: Solving for P(X=2, N=4), P(X=1), and P(N=4|X=1)

[cse63146]

Homework Statement

Suppose that the conditional distribution of X given that N = n is binomial (n, 1/2) and the distribution of N is uniform over {2,4,6}

a) Determine P(X=2, N = 4)
b) Determine P(X=1)
c) Determine P(N = 4| X =1)

Homework Equations

The Attempt at a Solution

the way I understood the question was that X|N~bin(n, 1/2)

for a) I did (4C2)(0.5)²(0.5)² = 0.375

I'm stuck for b). Any hints?

 

[HallsofIvy]

Homework Statement

Suppose that the conditional distribution of X given that N = n is binomial (n, 1/2) and the distribution of N is uniform over {2,4,6}

a) Determine P(X=2, N = 4)
b) Determine P(X=1)
c) Determine P(N = 4| X =1)

Homework Equations

The Attempt at a Solution

the way I understood the question was that X|N~bin(n, 1/2)

for a) I did (4C2)(0.5)²(0.5)² = 0.375

Yes, that is correct.

I'm stuck for b). Any hints?

Since N is uniform on 2, 4, 6, find P(X|N=2), P(X|N= 4), and P(X|N= 6). P(N= 2)= P(N= 4)= P(N= 6)= 1/3.

 

[Architeuthis Dux]

For b), you can use the formula for the binomial distribution to calculate P(X=1) as follows:

P(X=1) = (4C1)(0.5)1(0.5)3 = 4(0.5)(0.125) = 0.25

For c), you can use the conditional probability formula to calculate P(N=4|X=1) as follows:

P(N=4|X=1) = P(N=4 and X=1) / P(X=1)

Since N and X are independent, we can break this down into two separate probabilities:

P(N=4 and X=1) = P(N=4) * P(X=1) = (1/3) * (0.25) = 0.08333

P(X=1) = 0.25 (calculated in part b)

Therefore,

P(N=4|X=1) = 0.08333 / 0.25 = 0.33333