Let $X$ be the number of Heads in $10$ fair coin tosses.

(a) Find the conditional PMF of $X$, given that the first two tosses both land heads.

(b) Find the conditional PMF of $X$, given that at least two tosses land Heads.


(a) Let $Z$ be the number of heads in the first two tosses.

$ P(X=k|Z=2)={{10-2}\choose{k-2}}(\frac{1}{2})^{k-2}(\frac{1}{2})^{10-k} $

(b) We are interested in $P(X=k|X\ge2)$.

$ P(X=k|X\ge2)=\sum_{i=2}^{10}{{10-i}\choose{k-i}}(\frac{1}{2})^{k-i}(\frac{1}{2})^{10-k} $

Could someone please verify if my solution to the above problem is correct.




The first looks okay, though you really should indicate the support; $k\in\Bbb N{\cap}[2{;}10]$

The second, I'm not really sure what you are counting.   I would use Bayes' Rule.

$$\begin{align}\mathsf P(X=k\mid X\geq 2) &= \dfrac{\mathsf P(X=k)}{1-\mathsf P(X<2)} \\[1ex] & = \dfrac{\binom {10}k 2^{-10}\mathbf 1_{k\in\Bbb N{\cap}[2;10]}}{1-(\binom{10}{0}+\binom{10}1)2^{-10}} \\[1ex] & =\dfrac{\binom {10}k \mathbf 1_{k\in\Bbb N{\cap}[2{;}10]}}{2^{10}-11}\end{align}$$

  • $\begingroup$ Why do you have $2^{-10}$ in the numerator? $\endgroup$ – Matthew Graham Sep 24 '17 at 22:15

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