In his wardrobe, Fred has a total of ten pairs of gloves. He had to pack his suitcase before a business meeting, and he chooses eight gloves without looking at them. We assume that any set of eight gloves has an equal chance of being chosen.
I am told to calculate the likelihood that these 8 gloves do not contain any matching pairs, i.e. that no two (left and right) gloves are from the same pair.
This is what I came up with, that is, the probability of success for each choice:
$$\frac{20}{20}×\frac{18}{19}×\frac{16}{18}×…×\frac{6}{13}=\frac{384}{4199}≈0.09145$$
At first, I was a little confused by the wording but I believe this seems about right.
Is there an alternative way to get the desired probability, e.g. with $1-…$?
Thanks in advance for any feedback.
Best Answer
There is a more general formula for this.
Here you are asked that no pair is selected, but this formula will take care of any number of pairs selected
With $10$ be the number of pairs, and $k$ the number of pairs selected from $8$ gloves, the formula is
$\dfrac{\dbinom{10}{k}\dbinom{10-k}{8-2k}\cdot2^{8-2k}}{\dbinom{20}{8}}$
For the particular case for $k=0$, it simplifies to
$\dfrac{\dbinom{10}{0}\dbinom{10-0}{8-2\cdot0}\cdot2^{8-2\cdot0}}{\dbinom{20}{8}}$
$= \dfrac{\dbinom{10}{0}\dbinom{10}8 \cdot2^8}{\dbinom{20}{8}}$