Question

A closet contains 10 pairs of shoes, if 8 pairs of shoes are selected randomly, what is the probability that there is exactly one complete pair
A) \[\dfrac{{1792}}{{4199}}\]
B) \[\dfrac{{1782}}{{4199}}\]
C) \[\dfrac{{1772}}{{4199}}\]
D) \[\dfrac{{1762}}{{4199}}\]

Answer 1

Hint: We solve this question by first making all the possible combinations that we can make and then making the combinations which are required. There are a total 10 pairs which means that there are a total of 20 shoes. We have to select 1 pair from it and the rest different shoes so that no other pair is possible.

Complete step-by-step answer:
We know that there are 10 pairs which means that there are a total 20 shoes. We have to select 8 shoes from it which implies that we have total \[^{20}{C_8}\] ways through which we can make our selection.
Now, we have to find the probability of having exactly one complete pair. We know that, \[p\left( x \right) = \dfrac{m}{n}\], where p(x) is the probability of a given event, m is the number of outcomes in favor of x and n is the total number of outcomes.
Here, in the question, we have, n \[{ = ^{20}}{C_8}\]
We find m as follow
We have exactly one pair of shoes so we select it from 10 pairs and separate it which implies we have \[^{10}{C_1}\]ways to do so. Now we have 9 pairs of shoes from which we have to find 6 other shoes which do not form a pair. So the combinations will be \[^9{C_6}\]as we select 6 shoes from 9 pairs. Also we have to select one shoe from these 6 pairs which can be selected in \[{2^6}\]ways. This implies total ways to select 8 shoes with exactly 1 pair is, m, \[^{10}{C_1}\left( {^9{C_6}} \right)\left( {{2^6}} \right)\]
Substituting values of m and n we get,
\[p = \dfrac{{^{10}{C_1}{ \times ^9}{C_6} \times \left( {{2^6}} \right)}}{{^{20}{C_8}}}\]
Now, we know that \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]. So, substituting the values of combinations, we get,
\[ \Rightarrow p = \dfrac{{\left( {\dfrac{{10!}}{{1! \times 9!}}} \right) \times \left( {\dfrac{{9!}}{{6! \times 3!}}} \right) \times 64}}{{\left( {\dfrac{{20!}}{{12! \times 8!}}} \right)}}\]
Cancelling the common factors in numerator and denominator, we get,
\[ \Rightarrow p = \dfrac{{10 \times \left( {\dfrac{{9 \times 8 \times 7}}{{3!}}} \right) \times 64}}{{\left( {\dfrac{{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12!}}{{12! \times 8!}}} \right)}}\]
Simplifying the calculations, we get,
\[ \Rightarrow p = \dfrac{{10 \times \left( {\dfrac{{9 \times 8 \times 7}}{6}} \right) \times 64}}{{125970}}\]
\[ \Rightarrow p = \dfrac{{10 \times \left( {84} \right) \times 64}}{{125970}}\]
Cancelling common factors, we get,
\[ \Rightarrow p = \dfrac{{28 \times 64}}{{4199}}\]
Doing the multiplication, we get,
\[ \Rightarrow p = \dfrac{{1792}}{{4199}}\]
Hence, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: These questions deal with understanding of the concepts of combinations and probability. The basic formula of probability is \[p\left( x \right) = \dfrac{m}{n}\], where p(x) is the probability of given event, m is the number of outcomes in favor of x and n is the total number of outcomes. Also keep in mind the concepts of combinations for making selections. Take care while doing the calculations.