An urn contains balls numbered 1 through 6. Balls are repeatedly selected one at a time and with replacement. Let $X_3$ be the number of the selection on which the first 3 appears, and let $X_3$ be the number of the selection on which the first 5 appears. Let $p_{X_3|X_5}(x_3|x_5)$ be the conditional distribution of $X_3$, given that $X_5 = x_5$
Find $p_{x_3|x_5}(4|8)$ and $p_{x_3|x_5}(8|4)$
I tried using bayes theorem, but that did not work. Can someone help please?
I'll show you how to do the first one, then you can try the second one again.
Bayes' theorem is the right approach here. You want to compute the conditional probability $\mathbb{P}( X_3 = 4 \ | \ X_5 = 8).$ By Bayes' theorem, we have
$$\mathbb{P}( X_3 = 4 \ | \ X_5 = 8) = \frac{ \mathbb{P}(X_3 = 4 \ , \ X_5 = 8)}{\mathbb{P}(X_5=8)}$$
The denominator is easier to compute - each $X_i$ is just a Geometric random variable with parameter $p=1/6.$ The event $X_5 = 8$ is precisely the event where your first $7$ rolls are not $5$ and the $8$-th roll is $5.$ So $\mathbb{P}(X_5=8) = (5/6)^7 (1/6).$
The numerator is slightly harder. The event $X_3=4, X_5=8$ occurs precisely when the first $3$ rolls are not $3$ or $5$, the $4$-th roll is a $3$, the $5$-th, $6$-th and $7$-th rolls are not $5$, and the $8$-th roll is $5.$ So $\mathbb{P}(X_3 = 4 \ , \ X_5 = 8) = (4/6)^3 (1/6) (5/6)^3 (1/6).$ And then we get $$\mathbb{P}( X_3 = 4 \ | \ X_5 = 8) = \frac{4^3}{5^4}.$$
For the second part, do a very similar process of carefully writing down (via Bayes' theorem) the conditional probability you are required to calculate in terms of unconditional probabilities, and then understanding explicitly what events are occurring in those unconditional probabilities.