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.