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.
Best Answer
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.