I am solving Prob. 24 in Chapter - Random Variables and their distribution by Blitzstein and Hwang. Problem:
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.
Prob. 24, Random Variables and their distribution - Blitzstein and Hwang has already provided a solution but I have some doubts w.r.t. the solution.
@Quasar has provided solution to (a) as: (a) Let Z be the number of heads in the first two tosses.
$$P(X=k|Z=2)=\binom{10−2}{k−2}\frac{1}{2}^{k−2} \frac{1}{2}^{10−k}$$
But the way I'm doing, I'm getting something different:
E: be the event that first two tosses are H
$$P(E) = \frac{1}{2}.\frac{1}{2} = \frac{1}{4}$$
Desired probability is:
$$P(X=x|E) = \frac{P(X=x , \space E)}{P(E)}$$
And $P(X=x , \space E)$ is $P($No. of H in 10 flips are x and first two flips are H$)$. Since first two are H, so we consider remaining $x-2$ H in remaining $10-2$ flips. So,
$$P(X=x , \space E) = \binom{10-2}{x-2}(\frac{1}{2})^{x-2}(\frac{1}{2})^{10-x}$$
So, finally, $$P(X=x|E) = \frac{\binom{10-2}{x-2}(\frac{1}{2})^{x-2}(\frac{1}{2})^{10-x}}{\frac{1}{4}}$$ $$=\binom{10-2}{x-2}(\frac{1}{2})^{6}$$
Question: Why is he not dividing by 1/4 or am I doing it wrong?
Your $P(X=x, E)$ computation is actually the computation of $P(X=x|E)$. In your phrasing you say "Since first two are H, so we consider remaining $x−2$ H in remaining $10−2$ flips." The way you phrase this is saying that the first two are given to be H, and you have really calculated the conditional probability (correctly). Therefore no need to divide by $\tfrac14$.
P.S. good job on nicely detailed and formatted first post.