Two probability questions.

Question

I have two questions.

(1)

solution(1): Sample size $=|S|=12^{20}$

$11^{20}\rightarrow$ guarantee that one box is empty.

$10^{20}\rightarrow$ guarantee that two boxes are empty.

$9^{20}\rightarrow$ guarantee that three boxes are empty.

Then, $$P=\frac{10^{20}-9^{20}}{12^{20}}$$

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(2) a box contains 12pairs of shoes. If 8 shoes are rondomly selected, what is the probabilty that there willie exactly one complete pair?

Answer: $$\frac{12\binom{11}{6} 2^6}{\binom{20}{8}}$$

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I solved first question. But I am not sure the solution. Please check it. And I cannot understand the second question's answer. How is the answer written? Please explain it step by step. Thank you:)

Answer 1

There are $\binom{24}{8}$ equally likely ways to choose $8$ shoes.

Let us count the number of ways to have exactly one pair. The "type" of shoe we have two of can be chosen in $\binom{12}{1}$ ways. That leaves $6$ non-matching shoes to be chosen.

The types of shoe we have $1$ each of can be chosen in $\binom{11}{6}$ ways. And for every collection of $6$ types, we can choose the left shoe or the right shoe of each type, for a total of $2^6$. Thus the number of "favourables" is $\binom{12}{1}\binom{11}{6}2^6$.

The first answer is correct.

Answer 2

I think the answer to the first one is $\binom{28}{9}/\binom{31}{11}$.
Reason:
This is a partition problem since the balls are identical. Partitioning 20 identical balls among 10 boxes can be done in $\binom{29}{9}$. But we put one ball in the 3rd (fixed) box. And we are only left with 19 balls to be distributed among 10 boxes(boxes 3-12). Hence $\binom{28}{9}$.

Number of ways of distributing n identical objects among r groups