A fair die (with face numbered $1,\ \ldots\ ,6)$ is independently rolled repeatedly. Let $X$ denote the number of rolls till an even number is seen and let $Y$ denote the number of rolls till $3$ is seen. Evaluate $E(Y\ |\ X=2)$.
The way I've thought of it is, since $X=2,$ so on the first roll there must be an odd number. $P(Y=1\ |\ X=2)=\frac{1}{3} (3\ \mathrm{came\ on\ the\ 1^{st}\ roll})$ $P(Y=2\ |\ X=2)=\frac{2}{3}. 0\\ (3\ \mathrm{on\ the\ 2^{nd}\ roll,\ which\ contradicts\ with\ the\ fact\ that\ an\ even\ number\ came\ up\ on\ the\ 2^{nd}\ roll})$
For any other value of $Y$, on the $1^{st}$ roll an odd number other that $3$ came up (with probability $\frac{2}{3}$), on the $2^{nd}$ roll an even number came up (with probability $\frac{1}{2}$), on the $3^{rd}$ roll onwards either $3$ comes up (with probability $\frac{1}{6}$) or any number other than $3$ comes up (with probability $\frac{5}{6}$). Few of the cases are shown:
$P(Y=3\ |\ X=2)=\frac{2}{3}.\frac{1}{2}.\frac{1}{6}=\frac{1}{18}$
$P(Y=4\ |\ X=2)=\frac{2}{3}.\frac{1}{2}.\frac{5}{6}.\frac{1}{6}=\frac{1}{18}.\left(\frac{5}{6}\right)^{4-3}$
$P(Y=5\ |\ X=2)=\frac{1}{18}.\left(\frac{5}{6}\right)^{5-3}$
$\ldots$
$P(Y=n\ |\ X=2)=\frac{1}{18}.\left(\frac{5}{6}\right)^{n-3}$
So, $E(Y\ |\ X=2) = 1.\frac{1}{3} + 2.0 + 3.\frac{1}{18}+\ldots+n.\frac{1}{18}.\left(\frac{5}{6}\right)^{n-3}$
Is my approach and/or answer correct ? Can someone verify ?
$P\left(X=2\right)=\frac{3}{6}\times\frac{3}{6}=\frac{1}{4}$
$P\left(Y=1\wedge X=2\right)=\frac{1}{6}\times\frac{3}{6}=\frac{1}{12}$ hence $P\left(Y=1\mid X=2\right)=\frac{4}{12}=\frac{1}{3}$
$P\left(Y=2\wedge X=2\right)=0$ hence $P\left(Y=2\mid X=2\right)=0$
for $n\geq3$:
$P\left(Y=n\wedge X=2\right)=\frac{2}{6}\times\frac{3}{6}\times\left(\frac{5}{6}\right)^{n-3}\times\frac{1}{6}=\frac{1}{36}\times\left(\frac{5}{6}\right)^{n-3}$ hence $P\left(Y=n|X=2\right)=\frac{4}{36}\times\left(\frac{5}{6}\right)^{n-3}=\frac{1}{9}\times\left(\frac{5}{6}\right)^{n-3}$
This enables you to find $\mathbb{E}\left(Y\mid X=2\right)=\sum_{n=1}^{\infty}nP\left(Y=n\mid X=2\right)$
An alternative route:
Working unconditionally we have $\mathbb{E}Y=\frac{1}{6}\times1+\frac{5}{6}\times\left(1+\mathbb{E}Y\right)$ leading to $\mathbb{E}Y=6$.
Working under condition $X=2$ face $3$ will be seen at the first roll with probability $\frac{1}{3}$. If face $3$ does not appear at the first roll then from the third roll we can go on unconditionally. This because the condition does not affect the probabilities connected with these rolls.
This gives us equation: $\mathbb{E}\left(Y\mid X=2\right)=\frac{1}{3}\times1+\frac{2}{3}\times\left(2+\mathbb{E}Y\right)=\frac{17}{3}$