2
$\begingroup$

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.

Solution.

(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.

Thanks,

Quasar.

$\endgroup$
1
$\begingroup$

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.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.