probability question- expectation

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.

-
What work have you done towards solving this problem, and where are you having difficulties? – Dilip Sarwate Sep 16 '11 at 22:14

There are $\dbinom{10}{3}$ ways to choose $3$ of the $10$ chips. $\dbinom83$ of those ways result in getting $3$ of the $2$ dollar chips and therefore receiving $6$ dollars, so the probability of this happening is $$\frac{\binom83}{\binom{10}{3}}.$$

1. How many ways are there to get two $2$ dollar chips and one $5$ dollar chip? What is the probability that this will happen? How much will the person get in that case?
2. How many ways are there to get one $2$ dollar chip and two $5$ dollar chips? What is the probability that this will happen? How much will the person get in that case?
3. How many ways are there to get three $5$ dollar chips? What is the probability that this will happen? How much will the person get in that case?

The expected payoff is the weighted sum of the possible payoffs, each weighted by its probability. Once you’ve answered questions $(1)-(3)$, you’ll have all the information that you need to calculate this number.

-

Hints:

• What is the the sum of all the coins?

• What is $\dfrac{3}{10}$ of this?

-

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 $W$.

-

I think it's convenient to look at this problem as a simple binary tree with three steps. After each step you obtain either a 2\$or 5\$ chip, the only exception being outcome $<5,5,5>$ since there are only 2 5\$chips. Intuitively, the probability of the outcome, say,$5,5,2$is different to$2,5,5$, but in fact it's not, because$\frac{8 \cdot 7 \cdot 2}{10 \cdot 9 \cdot 8}=\frac{2 \cdot 8 \cdot 7}{10 \cdot 9 \cdot 8}$since multiplication is commutative. So all you are left to do is to find the number of$\textit{unique}$outcomes and multiply by their probability and value.$\mathbf{E}S=6 \cdot 1 \cdot \frac{8 \cdot 7 \cdot 6}{10 \cdot 9 \cdot 8} + 9 \cdot 3 \cdot \frac{8 \cdot 7 \cdot 2}{10 \cdot 9 \cdot 8} + 12 \Bigg( 2 \cdot \frac{2 \cdot 8 \cdot 1}{10 \cdot 9 \cdot 8} + \frac{2 \cdot 1}{10 \cdot 9} \Bigg)$The last term in the brackets is due to the fact that once ou'vr selected 2 5\$ chips, you are guaranteed to sample the 2\\$ one.

-