Probability - 12 pairs of shoes
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 ?
$\endgroup$ 42 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$I interpret the first part to mean that true pairs may be there, but "pseudo-pairs" are disallowed.
Thus $3$ cases arise: (i) $5 L$ or $5 R\;\;$ (ii) $1$ true pair + $3L$ or $3 R\;\;$ (iii) $2$ true pairs and $1 L$ or $1 R$ more
and $Pr = \dfrac{2\dbinom {12}5 +\dbinom{12}1\cdot2 \dbinom{11}3 +\dbinom{12}2\cdot2 \dbinom{10}1}{\dbinom{24}{5}}$
And just for completeness, for part two, with the numerator ruling out pair formations,
$Pr = \dfrac{24\cdot22\cdot20\cdot18\cdot16}{24\cdot23\cdot22\cdot21\cdot20}$
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