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Question: A box contains n pairs of shoes and 2r shoes are selected, (\(r < n\)). The probability that there i...

A box contains n pairs of shoes and 2r shoes are selected, (r<nr < n). The probability that there is exactly one pair is:
(A)n−1Cr−12nC2r\left( A \right)\dfrac{{{}^{n - 1}{C_{r - 1}}}}{{{}^{2n}{C_{2r}}}}
(B)nn−1Cr−12nC2r\left( B \right)n\dfrac{{{}^{n - 1}{C_{r - 1}}}}{{{}^{2n}{C_{2r}}}}
(C)(nC1n−1Cr−1)2r−12nC2r\left( C \right)\dfrac{{\left( {{}^n{C_1}{}^{n - 1}{C_{r - 1}}} \right){2^{r - 1}}}}{{{}^{2n}{C_{2r}}}}
(D)\left( D \right) None of these

Explanation

Solution

Hint – In this particular question use the concept that the number of ways to select r objects out of n objects is given as nCr{}^n{C_r} and use the property that in a pair of shoes there are two shoes, so use these concepts to reach the solution of the question.

Complete step by step solution:
Given data:
A box contains n pairs of shoes and 2r shoes are selected.
As in one pair of shoes there are two shoes.
So in n pairs of shoes there are 2n shoes.
Now we have selected 2r shoes out of n pairs of shoes.
So the number of ways in which we can choose 2r shoes out of 2n shoes = 2nC2r{}^{2n}{C_{2r}}
So the total number of outcomes = 2nC2r{}^{2n}{C_{2r}}.
Now we have to find the probability that there is exactly one pair.
So the number of ways to select one pair out of n pairs = nC1{}^n{C_1}
Now the remaining pairs of shoes = (n – 1)
As one pair of shoes is selected so we have to choose another (2r - 2) shoes.
So the number of pairs of shoes we have to choose (r – 1) but these pairs are not the matching pairs.
So this can be done in n−1Cr−1{}^{n - 1}{C_{r - 1}}
Now from these (r – 1) pairs of shoes the number of ways to select single unmatching shoes, this can be done in 2r−1{2^{r - 1}} ways.
So the favorable number of outcomes = (nC1n−1Cr−1)2r−1\left( {{}^n{C_1}{}^{n - 1}{C_{r - 1}}} \right){2^{r - 1}}
Now as we know that the probability (P) is the ratio of the favorable number of outcomes to the total number of outcomes.
Therefore, P=favorable number of outcomestotal number of outcomesP = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}
Now substitute the values we have,
Therefore, P=(nC1n−1Cr−1)2r−12nC2rP = \dfrac{{\left( {{}^n{C_1}{}^{n - 1}{C_{r - 1}}} \right){2^{r - 1}}}}{{{}^{2n}{C_{2r}}}}
So this is the required probability.
Hence option (C) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the property of combinations and that the probability is the ratio of the favorable number of outcomes to the total number of outcomes, so first find out the total number of outcomes then the favorable number of outcomes as above then use probability formula we will get the required answer.