Question

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) $\dfrac{{{}^{20}{C_1} \times \left( {{}^{38}{C_2} - 19} \right)}}{{{}^{40}{C_4}}}$
C) $\dfrac{{{}^{20}{C_1} \times {}^{38}{C_2}}}{{{}^{40}{C_4}}}$
D) $\dfrac{{{}^{20}{C_1} \times {}^{19}{C_2}}}{{{}^{40}{C_4}}}$

Answer 1

In this problem there are 20 pairs of shoes. That is a total of 40 shoes. From these 40 we have to select 4 shoes. But the condition is these four shoes should have exactly one pair. So the combination is one pair and two shoes separately.

Complete step by step solution:
Given that there are 20 pairs of shoes in a closet.
But four shoes are selected at random.
And this selection has only one pair.
So probability would be,
$\dfrac{{{}^{20}{C_1}}}{{{}^{40}{C_4}}}$
But the remaining two shoes are not in pairs. So there can be any two shoes from the remaining 38 shoes.
Now the ratio changes to
$\dfrac{{{}^{20}{C_1} \times {}^{38}{C_2}}}{{{}^{40}{C_4}}}$
But if one pair and two shoes are already selected then 1 shoe from the remaining 19 pairs should be removed from selection.
So final ratio of probability will become
$\dfrac{{{}^{20}{C_1} \times \left( {{}^{38}{C_2} - 19} \right)}}{{{}^{40}{C_4}}}$

Hence B is the correct option.

Note:
Now why not remaining options is the question to be solved.
For option A: it only means that we are selecting 4 shoes and one of them is a pair. But there is no significance of remaining two shoes. So we eliminate it.
For option C: it tells us about one pair and two shoes but no idea of remaining 19 single shoes.
For option D: It tells us about one pair of shoes but it is taking one more pair from the remaining 19 pairs. And that is not the given condition.
So we selected option B as the correct option.