There are $12$ pairs of shoes in a cupboard. $4$ are drawn at random. What is the probability that there is at least one pair?
My first attempt: If we chose a pair at first and then draw any two at random from the rest, then there will be at least one pair.
We can choose one pair in ${12}\choose 4$ ways and chose any $2$ from the rest in ${22}\choose 2$ ways.
Therefore, the required probability= $\frac{{12\choose 4} \times {22\choose 2}}{{24\choose 4}}$ = $\frac{6}{23} =\frac{42}{161}$
But the given answer is $\frac{41}{161}$.
Another attempt: Each of the 4 shoes we choose, will come from one of the pairs. We can choose the four pairs in ${12\choose 4}$ ways and can select a shoe from each of the pairs in $2$ ways so that no pair is obtained. Therefore, required probability =$1-$ $\frac{{12\choose 4} \times 2^4}{{24\choose 4}}$ = $\frac{41}{161}$
What is wrong with the first attempt?
Best Answer
Calculate $1$ minus the probability of the complementary event:
The number of ways to choose $4$ out of $24$ shoes is:
The number of ways to choose $4$ out of $24$ shoes with no pairs is:
So the probability of choosing $4$ out of $24$ shoes with at least one pair is:
$$1-\frac{24\cdot22\cdot20\cdot18}{24\cdot23\cdot22\cdot21}$$
Please note that I've essentially taken into account the order of the shoes.
If I chose not to take it into account, then I would need to divide each result by $4!$.
But since this factor appears in both the numerator and the denominator, I can ignore it.