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

AI Thread Summary
The discussion revolves around solving a binomial distribution problem where X given N follows a binomial distribution with parameters (n, 1/2) and N is uniformly distributed over {2, 4, 6}. For part (a), the calculation for P(X=2, N=4) is confirmed as correct at 0.375. Participants suggest that for part (b), to find P(X=1), one should calculate P(X|N=2), P(X|N=4), and P(X|N=6), noting that each has a probability of 1/3 due to the uniform distribution of N. The discussion highlights the need for understanding conditional probabilities in the context of binomial distributions. The thread emphasizes the importance of breaking down the problem into manageable parts for clarity in solving.
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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?
 
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cse63146 said:

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.
 
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