# Combinations/Probability Question

A high school lottery uses two sets of numbered balls. One set consists of ten white balls numbered 1-10 and the second set contains twenty blue balls numbered 1-20. To play, you select two white balls and two blue balls.

(a) How many different outcomes are possible?

(b) Your lottery ticket consists of four numbers: two white numbers, each between 1-10 inclusive, and two blue numbers, each between 1 and 20, inclusive. What is the probability that your lottery ticket contains exactly one matching one number and two matching blue numbers?

(a) This part is easy: $$\binom{10}{2}\binom{20}{2}=8,550$$

(b) This part I don't understand how to get the right answer. For one of the wrong white balls, you have 8 choices because you have to subtract the 2 winning numbers. This is represented as $$\binom{10}{8}$$. There is only one possibility for the winning numbers WBB, so I didn't think I had to do anything so the solution would be:

$$\cfrac{\binom{8}{1}}{\binom{10}{2}\binom{20}{2}}=0.0009356$$, which is wrong.

I thought about putting in other terms like $$\binom{20}{2}$$, but the solution becomes too large... I'm not sure what I'm missing. The book solution is $$0.00187$$.

Any help is appreciated. Thanks in advance.

Among the $10$ white balls, there are two "good" ones, the winning ones, and $8$ "bad" ones, the ones that did not win. Your one good white number can be chosen in $\binom{2}{1}$ ways. For each such choice, the bad white can be chosen in $\binom{8}{1}$ ways, for a total of $\binom{2}{1}\binom{8}{1}$ "favourables." Now divide by the correct total $\binom{10}{2}\binom{20}{2}$ of equally likely possibilities.
• Thanks Andre! I see that $\cfrac{2*8}{8,550}=0.00187$, the correct answer. But if we have to accommodate for both of the white balls to derive the solution, how come we don't do the same for the blue balls. The two winning blue balls could be chosen in $\binom{20}{2}$ ways. Why is this not used? Commented May 3, 2013 at 16:04
• user1527227 To answer your question in the comment about the ways two winning blue numbers can be chosen, I think @AndréNicolas meant there are only two winning blue numbers, and your selection must contain both of them. Hence $\binom{2}{2} = 1$ way to meet that condition. I.e., no choice involved with respect to the winning blue numbers. Numerator: $\binom 21 \binom 81 \binom 22 = 16$ Commented May 3, 2013 at 16:25
• @user1527227: Note the typo correction by amWhy in my comment. The "full" expression for the numerator is $\binom{2}{1}\binom{8}{1}\binom{2}{2}\binom{18}{0}$, but I left out the last two terms when giving the answer, since they are both equal to $1$. Commented May 3, 2013 at 16:29