# A bag contains five counters marked 1 to 5, expected value and variance

A bag contains five counters, marked with the numbers 1 to 5. A person draws two counters from the bag and is to obtain an amount in euro equal to the product of the two numbers shown on the counters. Find the expected value and variance of the amount the person will obtain.

I have tried plugging in 1 to 5 in expected values but I am way off getting 2.93333333 For the Variance using the values found for expected value I am getting 2.004444444

How should I go about approaching this question

You haven't told us if the counters are replaced or not.

That is, do we draw 1 counter, put it back, and draw a second one, such that we can draw the same one twice?

Or do we draw two without replacement, such that we cannot draw the same one twice?

This will make a big difference.

I get an expectation value of €6 if there is replacement of counters.

Either way, to go about calculating it, you have to calculate the profit to be made on each outcome, and then sum the probability of each outcome multiplied by it's profit.

So, for example, if we assume no replacement, then we can have $$5 \times 5 = 25$$ possible outcomes from drawing two counters from a bag, agreed?

Then all things being equal, there is a probability of $\frac{1}{25}$ of each of the outcomes happening.

Say you draw a 5 and then a 4, your profit would be $$€5 + €4 = €9$$ with probability of $$2 \times \frac{1}{25}$$ of happening.

This gives a contribution of $$€ 9 \times \frac{2}{25} = € \frac{18}{25}$$ to your expected profit.

Then repeating for each possible outcome and summing will give the total expected profit, which I work out to be €6.

Now, in the case of replacement of the counters, the two games (draw a counter, win money) are unrelated. That is, they are two memoryless events, with independent probabilities.

In this case, we can calculate the expected profit of drawing a disk, and multiply by two, since we play the game twice.

This is far less tedious! We get, since each counter is equally likely to be drawn, $$E(profit) = 2 \times [€5 \times \frac{1}{5} + €4 \times \frac{1}{5} + €3 \times \frac{1}{5} + €2 \times \frac{1}{5} + €1 \times \frac{1}{5}] = 2 \times [€ 3] = € 6$$

Does this help?

The variance follows by similar method, using the variance equation of course. I get $€ 2$