There are 12 pairs of shoes in a closet. Five shoes are picked at random.
(a)What is the probability there is no "pseudo-pair"(i.e., one left and one right shoe)?
(b) What is the probability that there is no pair?
I calculated the sample size to be |S| = (24 choose 2)
For (a) the event A where there is no pseudo pair |A| = (12 choose 2) so that P(A) = |A|/|S| would this be correct or am I missing something ?
And for part (b) wouldn't the answer be the same as in part (a) ? Or perhaps I'm misinterpreting the question ?
Best Answer
There are $\binom{24}{5}$ equally likely possible choices of $5$ shoes. We want the probability that all the shoes are left shoes or all right shoes, There are $\binom{12}{5}$ ways to choose all left, and the same number of ways to choose all right. So the required probability is $\frac{2\binom{12}{5}}{\binom{24}{5}}$.
As to the probability of no matching pair of shoes, we count the number of ways to choose $5$ shoes, no two of which form a pair. There are $\binom{12}{5}$ ways of choosing $5$ shoe "types." For each such choice, there are $2^5$ ways to choose the actual shoes, for a total of $\binom{12}{5}2^5$. For the probability, divide by $\binom{24}{5}$.
Another way: We solve the second problem. A similar but easier argument deals with the first problem. Imagine picking the shoes one at a time.
Whatever shoe we picked first, the probability the second does not match is $\frac{22}{23}$. Given that there was no match in the second pick, there are $22$ shoes left, of which $20$ don't match either of the first two. So the probability of no match after the third pick is $\frac{22}{23}\cdot \frac{20}{22}$. Given there was no match among the first $3$, the probability of no match on the fourth is $\frac{18}{21}$. Continue, it's almost over.