Let $X$ be the number of tosses of a fair coin required to get the first head. If $Y | X = n$ is distributed as Binomial$(n, \frac{1}{2})$, then what is $P(Y = 1)$?

(A) 4/9 
(B) 1/4 
(C) 1/3 
(D) 5/9

What I understand so far is that $X$ follows geometric distribution. The other part, I couldn't relate. Any help would be appreciated.


A typical way to solve it would be applying Total Probability Law: $$P(Y=1)=\sum_{n=1}^\infty P(Y=1|X=n)P(X=n)=\sum_{n=1}^\infty n\left(\frac{1}{4}\right)^n$$

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  • $\begingroup$ Why multiply by n in the summation? $\endgroup$ – Muskaan Madan Sep 16 at 12:49
  • 1
    $\begingroup$ Try to write $P(Y=1|X=n)$ $\endgroup$ – gunes Sep 16 at 12:50
  • $\begingroup$ That will be $(1/2)^n$. Did I write correctly? $\endgroup$ – Muskaan Madan Sep 16 at 12:52
  • $\begingroup$ No, you forget ${n\choose 1}$, check en.wikipedia.org/wiki/Binomial_distribution $\endgroup$ – gunes Sep 16 at 13:14
  • $\begingroup$ So, the answer would be 1/4? $\endgroup$ – Muskaan Madan Sep 16 at 15:33

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