I was studying the concept of random variables when I encountered an exercise problem (from Introduction to Probability (2e) – Blitzstein & Hwang). I noticed that somebody had already asked a question regarding the exact same exercise problem here. However, it didn't address the question I had, which prompted me to ask my own.
Question:
Let $X$ be the number of Heads in 10 fair coin tosses. Find:
(a) the conditional PMF of $X$, given that the first two flips landed Heads, and
(b) the conditional PMF of $X$, given that at least two flips land Heads.
I derived solutions to both, but had a question regarding the particular solution to (b).
My Solution:
$$P(X = k\ |\ X \ge 2)\ =\ \frac{P(X = k,\ X \ge 2)}{P(X \ge 2)}$$
$$=\ \frac{P(X = k)}{P(X \ge 2)}$$
$$=\ \frac{P(X = k)}{1\ -\ (P(X = 0)\ +\ P(X = 1))}$$
$$=\ \frac{\binom{10}{k}(\frac{1}{2})^{10}}{\frac{1013}{1024}}$$
$$=\ \frac{\binom{10}{k}}{1013}$$
(If my solution is incorrect, please do let me know.)
My specific question has to do with the first line that I wrote:
How do we come to the conclusion that $P(X=k,\ X \ge 2)\ =\ P(X=k)$?
My initial thought is that this line of reasoning may come from the fact that we'd be dividing the cases to where ($k = 0,\ 1$) and ($k \ge 2$), so it wouldn't matter which particular probability we're using. Is that correct?
Any feedback is appreciated. Thank you.
Best Answer
As you write $P(X=k, X \ge 2)$, I am going to assume that $k \ge 2$, otherwise the probability is $0$.
Now, if $k \ge 2$, $Pr(X =k , X \ge 2)=Pr(X =k)$ since $X=k$ have already imply that $X \ge 2$.
$$Pr(X=k, X \ge 2) = \begin{cases} 0 &, k < 2 \\ Pr(X=k) &, k \ge 2. \end{cases}$$