
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
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.
\[P(A) = \dfrac{{favourable\left( {event} \right)}}{{total\left( {event} \right)}}\].
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which one of the following is a true fish A Jellyfish class 12 biology CBSE

Which are the Top 10 Largest Countries of the World?

Why is insulin not administered orally to a diabetic class 12 biology CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

The total number of isomers considering both the structural class 12 chemistry CBSE
