Question

a

Answer 1

Use the conditional probability formula:
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
In this case,
\[P(X\ge4|X\ge2)=\frac{P(X\ge2\text{ and }X\ge4)}{P(X\ge2)}=\frac{P(X\ge4)}{P(X\ge2)}\]

Answer 2

The reason you can simplify the numerator like that is because of \(X\ge4\), then it is automatically true that \(X\ge2\).

Answer 3

Yep, that's correct. Compute that and you have your answer.

Answer 4

If you have a solid understanding of conditional probability, it shouldn't be an issue. You're finding the probability that a success is attained on *at least* the 4th try, given that a success is attained on *at least* the 4th try. (I emphasize the "at least" part because it doesn't make sense to have a success on the first try, for example, given that there's a success on the second; these two events would be necessarily mutually exclusive.)

Answer 5

Yeah, sorry, it should say "on at least the 4th, given a success on at least the 2nd."

Answer 6

It wasn't *necessarily* achieved on the 2nd try, though. A success occurs on *at least* the 2nd try, which means it could have happened on the 2nd, 3rd, 4th, etc.

Answer 7

Basically, you're finding the probability that a success happens on the 4th or later try, given that it didn't happen on the first.