Conditional Expected Number of Balls Left in the Urn

Question

Question: An urn contains three balls. The number of red balls in the urn is $0,1,2$ or $3$ equally likely. A ball is drawn from the urn at random. Given that the ball drawn is red, what is the conditional expected number of red balls left in the urn?

Comment: Poorly worded question, but with it were two equations for utilization:

1. Conditional expectation. $E(X\mid Y=y)$ denotes the conditional expectation of $X$ given $Y=y$
2. Law of Iterated Expectations. $E(X\mid Y=y)Py(y)$

Thank you!!

Answer 1

You seek $\mathsf E(X\mid Y=1)$, where $X$ is the count of red balls put into the urn, and $Y$ is the indicator that the ball drawn out was red.

You know what, for any $k\in\{0,1,2,3\}$ are $\mathsf P(X=k)$ and $\mathsf P(Y=1\mid X=k)$.   The probability that $k$ balls were put in the urn, and the probability that a red ball was extracted given that there were $k$ balls in the urn.

$$\begin{split}\mathsf P(X=k)&~=~\tfrac 14\mathbf 1_{x\in\{0,1,2,3\}}\\ \mathsf P(Y=1\mid X=k)&~=~\tfrac k3\mathbf 1_{k\in\{0,1,2,3\}}\end{split} $$

So you can find out $\mathsf P(Y=1,X=k)$, $\mathsf P(Y=1)$, and hence $\mathsf P(X=k\mid Y=1)$.

$$\begin{split}\mathsf P(Y=1,X=k)&~=~\phantom{\tfrac k{12}\mathbf 1_{k\in\{0,1,2,3\}}}\\\mathsf P(Y=1)&~=~\phantom{\tfrac 12}\\ \mathsf P(X=k\mid Y=1)&~=~\phantom{\tfrac k6\mathbf 1_{k\in\{0,1,2,3\}}}\end{split}$$

From which finding what is $\mathsf E(X\mid Y=1)$, is easy.

$$\mathsf E(X\mid Y=1)~=~\phantom{\tfrac 73}$$

Answer 2

Write $X$ for the number of red balls. The probability $P(X=i\wedge R)$, where $R$ is the event of drawing a red ball, is $\frac 14\times \frac i3$ (having $i$ red balls, then choosing one of them). The total probability of seeing a red ball is $\sum_{i=0}P(X_i\wedge R)$ (sanity check: this should be $\frac12$ by symmetry).

The conditional probability satisfies $$P(X=i\mid R)=\frac{P(X=i\wedge R)}{P(R)}$$ for $i=0,1,2,3$. You should be able to finish from here.