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Question: There are 20 pairs of shoes in a closet. Four shoes are selected at random. The probability that the...

There are 20 pairs of shoes in a closet. Four shoes are selected at random. The probability that there is exactly one pair is
A. 20C140C4\dfrac{{^{20}{C_1}}}{{^{40}{C_4}}}
B. 20C1(38C2−19)40C4\dfrac{{^{20}{C_1}\left( {^{38}{C_2} - 19} \right)}}{{^{40}{C_4}}}
C. 20C1(38C2)40C4\dfrac{{^{20}{C_1}\left( {^{38}{C_2}} \right)}}{{^{40}{C_4}}}
D. 20C1(19C2)40C4\dfrac{{^{20}{C_1}\left( {^{19}{C_2}} \right)}}{{^{40}{C_4}}}

Explanation

Solution

First do the selection of random selection of four shoes out of 20 pairs of shoes in a closet. And then apply the condition to select exactly one pair and the remaining two shoes are to be selected in such a way that they are from the different pairs. The ways of selecting r things from n total things are ncr^n{c_r}. And the probability of an event is given as
P(A)=favourable(event)total(event)P(A) = \dfrac{{favourable\left( {event} \right)}}{{total\left( {event} \right)}}.

Complete step by step answer:

As per the given there 2020 pairs of shoes in a closet.
The ways of selecting r things from n total things are ncr^n{c_r}
Hence, to select four random shoes among in total 4040 shoes is given as =40c4{ = ^{40}}{c_4}
And now we have to select four shoes in such a way that among four shoes there should be one pair and rest two shouldn’t form a pair as,
Hence, to select a pair among total pair of shoes is given as =20c2{ = ^{20}}{c_2}
And now there are total 1919 pair of shoes from which we have to select two shoes from different pair as
=(38c2−19)= \left( {^{38}{c_2} - 19} \right)
This means we are doing the selection of two shoes from 3838 shoes and eliminating one pair of shoes of each pair.
Hence, the total ways to select four shoes from pair of 2020 shoes such that only one pair is formed and remaining are from different pair as,
=20c1(38c2−19){ = ^{20}}{c_1}\left( {^{38}{c_2} - 19} \right)
Hence, probability can be calculated as P(A)=favourable(event)total(event)P(A) = \dfrac{{favourable\left( {event} \right)}}{{total\left( {event} \right)}}
So, the above probability can be given by
P(A)=20c1(38c2−19)40c4P(A) = \dfrac{{^{20}{c_1}\left( {^{38}{c_2} - 19} \right)}}{{^{40}{c_4}}}
Hence, option (B) is correct answer.

Note: Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. ... This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. Remember that combinations are a way to calculate the total outcomes of an event where the order of the outcomes does not matter. To calculate combinations, we will use the formula ncr=n!r!(n−r)!^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} where n represents the number of items, and r represents the number of items being chosen at a time. In the above problem don’t forget to eliminate 1 shoe from each pair as we need to select only one shoe from 19 pairs.