Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

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. \[\dfrac{{^{20}{C_1}}}{{^{40}{C_4}}}\]
B. \[\dfrac{{^{20}{C_1}\left( {^{38}{C_2} - 19} \right)}}{{^{40}{C_4}}}\]
C. \[\dfrac{{^{20}{C_1}\left( {^{38}{C_2}} \right)}}{{^{40}{C_4}}}\]
D. \[\dfrac{{^{20}{C_1}\left( {^{19}{C_2}} \right)}}{{^{40}{C_4}}}\]

Answer
VerifiedVerified
510.9k+ views
Hint: 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 \[^n{c_r}\]. And the probability of an event is given as
\[P(A) = \dfrac{{favourable\left( {event} \right)}}{{total\left( {event} \right)}}\].

Complete step by step answer:

As per the given there \[20\] pairs of shoes in a closet.
The ways of selecting r things from n total things are \[^n{c_r}\]
Hence, to select four random shoes among in total \[40\] shoes is given as \[{ = ^{40}}{c_4}\]
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 \[{ = ^{20}}{c_2}\]
And now there are total \[19\] pair of shoes from which we have to select two shoes from different pair as
\[ = \left( {^{38}{c_2} - 19} \right)\]
This means we are doing the selection of two shoes from \[38\] shoes and eliminating one pair of shoes of each pair.
Hence, the total ways to select four shoes from pair of \[20\] shoes such that only one pair is formed and remaining are from different pair as,
\[{ = ^{20}}{c_1}\left( {^{38}{c_2} - 19} \right)\]
Hence, probability can be calculated as \[P(A) = \dfrac{{favourable\left( {event} \right)}}{{total\left( {event} \right)}}\]
So, the above probability can be given by
\[P(A) = \dfrac{{^{20}{c_1}\left( {^{38}{c_2} - 19} \right)}}{{^{40}{c_4}}}\]
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 \[^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] 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.