Question

Suppose r.v. X takes values 1, 2, 3, 4. If its p.m.f. is given in the following table, what is the
conditional probability P(X = 3|2 ≤ X ≤ 4)?

x | 1 2 3 4
p(x)| .1k .2 .5k k

Answer 1

\[P(X=3~\bigg|~2\le X\le4)=\frac{P(X=3~\wedge~2\le X\le 4)}{P(2\le X\le 4)}=\frac{P(X=3)}{P(2\le X\le4)}\]
\(P(X=3)\) is given, and \(P(2\le X\le4)=P(X=2)+P(X=3)+P(X=4)\), which are all given.

However, an issue I'm having with this pmf is that the sum of probabilities does not add up to 1. Are you sure the table is correct?