A box contains 5 red and 5 blue balls. Two balls are withdrawn randomly. If they are the same color, then you win NT$11 dollars. However, if they are different colors, then you win –NT$ 10 dollars (that is, you lose NT$ 10 dollars). Calculate
(a) the expected value of the money you win.
(b) the variance of the money you win.








I will be Glad if you can help me !

First, you need to put up a probability distribution function (PDF).

Let X be the amount of money you win.


\begin{array}{|c|c|c|} \hline
X & -10 & \hspace{4pt} 11 \hspace{4pt} \\ \hline
P(X=x) & & \\ \hline \end{array}

Now you need to fill in the table. What is the probability of getting -NT$ 10?
That is the probability that you pick a red then a blue, or a blue then a red. Can you find this?

Then, put the probability in the table under -10

Do the same with the next one, gaining NT$ 11. (That is P(blue, blue or red red))

Once you filled in the table, you can go to the first part.

Expected value is obtained by adding XP(X=x). That is, multiply the probability with its event. Say you had 1/2 for the event you lose NT$ 10, and half for gaining NT$ 11, then the expected value will be

E(X) = (0.5 x -10) + (0.5 x 11) = -5 + 5.5 = NT$ 0.5

Variance is obtained by:

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

E(X^2) is taking first the square of the event and multiplying this with the corresponding probability.
E^2(X) is just the square of your expected value.

Post what you get! :)