A die is rolled repeatily. Let $X$ be the random variable that denotes the number of rolls to get a 4 and $Y$ be the random variable that denotes the number of rolls to get a 1. What is $E[X|Y=7]?
My thoughts were $\dfrac{1}{\dfrac{1}{6}} + 7$ since the expected value for rolling a 4 is 6 and we are given that we rolled 7 times (but we know on the 7th roll we did not get a 4)) but I know the answer is not right. Since we must factor in the probabilites of rolling a 4 in the first 6 rolls. How do I do this?
Answer
Hint: Note that $X$ is a geometric random variable. $Y=7$ implies that rolls one through to 6 was not a $1$. So we can consider two cases: $X \le 6$ and $X\gt 7$
By definition $$\begin{align*} E(X \, | \, Y = 7) & = \sum_{k=1}^{\infty} \, k \,P(X = k \, | \, Y = 7)\\
& = E(X \, | \, Y = 7, X\lt 7) \cdot P(X \le 6 \, | \, Y = 7) \\
&\,\,\,\,\,\,\,\,\,\,\,+ E(X \, | \, Y = 7, X\gt 7) \cdot P(X \gt 7 \, | \, Y = 7)
\end{align*}$$
Can you take it from here?
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