Question

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if they are different colors, then you win -$1.00. (This is, you lose $1.00.) Calculate (a) the expected value of the amount you win; (b) the variance of the amount you win.

Answer 1

Answer:

a) The expected value is $\frac{-1}{15}$

b) The variance is  $\frac{49}{45}$

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$.

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is $\binom{5}{2}$. Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is $\binom{5}{1}\cdot \binom{5}{1}$. So, we define the following probabilities.

Probability of winning: $\frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}$

Probability of losing $\frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}$

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing =$1.10\cdot \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}$

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing =$1.10^2\cdot \frac{4}{9}+\frac{5}{9}=\frac{82}{75}$

We will use the following formula

$Var(X) = E(X^2)-E(X)^2$

Thus

Var(X) = $\frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}$