law of total probability and conditiona probability exercise.

Question

Exercise:

Let $X$ be an uniform discrete r.v. with four possible values: 1, 2, 3, 4. Let $Y$ be an exponential variable whose parameter is the value taken by $X$. So, if $X = 3$, $Y$ is Exp (3). Determine:

The probability $P(Y≤ 1/2 | X>2)$

The solution gives $P(Y≤ 1/2 | X>2) = 1/2 P(Y≤ 1/2|X=3) + 1/2 P(Y≤ 1/2|X=4)$

My first attempt at solving the question was:

$P(Y≤ 1/2 | X>2) = \dfrac{ P(Y≤ 1/2 \cap X> 2) }{P(X>2)} = \dfrac{ P((Y≤ 1/2 \cap X> 2) \cap X=3)+P((Y≤ 1/2 \cap X> 2) \cap X=4) }{P(X>2)} = \dfrac{ P(Y≤ 1/2 \cap X=3)+P(Y≤ 1/2 \cap X=4) }{P(X>2)} = \dfrac{ P(X=3)P(Y≤ 1/2 | X=3)+P(X=4)P(Y≤ 1/2 | X=4) }{P(X>2)} $

But this is incorrect, why?

Is the solution applying the law of total probability to a conditional probability? How is this done? Could I see the formula used?

Answer 1

Your work seems correct to me. Note also that $\mathbb P(X=3)=\mathbb P(X=4)=1/4$ and $$\mathbb P(X>2)=\mathbb P(X=3)+\mathbb P(X=4)=\frac12,$$ so...

Answer 2

also note that $$\frac{P(X=3)}{P(X>2)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$$

So this would give you answer above. Just remember law of total probability just says the probability of a given event is a weighted average sum of all possible conditional probabilities.