# [Math] probability question- expectation

probability

A bowl contains 10 chips , of which 8 are marked 2 dollars each and 2 are marked 5 dollars each. Let a person choose at random and without replacement, 3 from this bowl. If the person is to receive the sum of the resulting amounts , find the expectation.

The hard way: One of several things can happen.

Case (a): We get three $2$'s;

Case (b): We get two $2$'s and a $5$;

Case (c): We get one $2$ and two $5$'s.

Let us find the probability of each. There are $\binom{10}{3}$ ways of choosing three chips, all equally likely.

Case (a): There are $\binom{8}{3}\binom{2}{0}$ ways of choosing three $2$'s and zero $5$'s. So the probability of (a) is $$\frac{\binom{8}{3}\binom{2}{0}}{\binom{10}{3}}.$$

Case (b): There are $\binom{8}{2}\binom{2}{1}$ ways to get two $2$'s and one $5$. So the probability of (b) is $$\frac{\binom{8}{2}\binom{2}{1}}{\binom{10}{3}}.$$

Case (c): Similarly, the probability of (c) is $$\frac{\binom{8}{1}\binom{2}{2}}{\binom{10}{3}}.$$

In Case (a) we get $6$ bucks, in Case (b) we get $9$, and in Case (c) we get $12$. So the expectation is $$6\frac{\binom{8}{3}\binom{2}{0}}{\binom{10}{3}}+9\frac{\binom{8}{2}\binom{2}{1}}{\binom{10}{3}}+12\frac{\binom{8}{1}\binom{2}{2}}{\binom{10}{3}}.$$

Finally, compute. We have $\binom{10}{3}=120$, $\binom{8}{3}\binom{2}{0}=56$, $\binom{8}{2}\binom{2}{1}=56$, and $\binom{8}{1}\binom{2}{2}=8$. So the expectation is $$\frac{6\cdot 56+9\cdot56+12\cdot 8 }{120}.$$

This turns out to be $936/120$, which simplifies to $39/5$, that is, $7.80$ dollars.

An easier way: Draw the chips one at a time. Let $X_1$ be the amount we get from the first chip, $X_2$ the amount we get from the second, and $X_3$ the amount we get from the third. Then our total winnings $W$ are given by $$W=X_1+X_2+X_3.$$ Thus $$E(W)=E(X_1+ X_2+X_3)=E(X_1)+E(X_2)+E(X_3).$$ (The expectation of a sum of random variables is always the sum of the expectations.) Let us find $E(X_1)$. Se get $2$ bucks with probability $8/10$, $5$ bucks with probability $2/10$, so $$E(X_1)=2\cdot\frac{8}{10}+5\cdot\frac{2}{10}=\frac{13}{5}.$$ By symmetry, $E(X_2)=E(X_3)=13/5$. It follows that $E(W)=39/5$.

Comment: Let $W$ be our winnings. In our first approach, we found $E(W)$ by first finding the distribution of the random variable $W$. This was slightly unpleasant but feasible. However, the second approach can help us find the expectations of certain random variables $W$ even when it is not feasible to find the probability distribution of $W$.