Exercise 1-28 A high school lottery uses two sets of numbered balls...

Question

Exercise 1-28 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 and 10 inclusive, and two blue numbers, each between 1 and 20, inclusive. What is the probability that your lottery ticket contains exactly one matching white number and two matching blue numbers?

#1) is easy, it’s 10C2 x 20C2 = 8550. I get the concept behind these. However, in #2) the answer is 0.00187 and the math is apparently (2C1x8C1x2C2x18C0)/8550 but what’s the reasoning behind this? Help this make sense to me. I got 0.001053 by using this math: 9/45 * 1/190. The reasoning behind this is that if we want at least 1 out of 10 white balls to match our lottery ticket, well each ball repeats 9 times out of all the 45 combinations. And if we want 2 out of 20 blue balls to match well the chances of that occurring is 1/190, out of all 190 combinations, yet it’s the wrong answer.

Answer 1

As you correctly observed, there are $\binom{10}{2}$ ways to select the numbers on the white balls and $\binom{20}{2}$ ways to select the numbers on the blue balls, so there are $$\binom{10}{2}\binom{20}{2}$$ outcomes in the sample space.

You wish to find the probability of matching one white ball and two blue balls with the two white numbers and two blue numbers on the lottery ticket. There is one way to match both blue numbers on the lottery ticket: You must select the two winning blue numbers and none of the $18$ losing blue numbers. To match exactly one white number on the lottery ticket, you must select one of the two winning white numbers and one of the eight losing white numbers. Hence, the probability of matching exactly one white number and two blue numbers on the lottery ticket is $$\frac{\dbinom{2}{1}\dbinom{8}{1}\dbinom{2}{2}\dbinom{18}{0}}{\dbinom{10}{2}\dbinom{20}{2}}$$

Answer 2

Well, since you have edit in working/doubts, I have to say that the question is not at all clear, e.g. matching what ?

However, reverse engineering from the book answer, it seems that $2$ white and $2$ blue are the winning numbers to be matched, so the explanation of the second part is

Pr = (Draw $1$ winning white, $1$ non-winning white, $2$ winning blues, $0$ non-winning blues)/(total ways of drawing)

$ =\dfrac{\dbinom21\dbinom81\dbinom22\dbinom{18}0}{\dbinom{10}2\dbinom{20}2}$