Question

ack has a collection of 10 pairs of gloves in his wardrobe. Before a business trip, he has to pack his luggage, and he selects 8 gloves, without looking at them. We assume that any set of 8 gloves is equally likely to be chosen. Find the probability that these 8 gloves do not include any matching pair of gloves, that is, that there are no two (left and right) gloves, coming from the same pair.

Answer 1

Answer:

$\frac{{10 \choose 8}2^8}{{20 \choose 8}}\approx 0.091$

Step-by-step explanation:

We can think of the 10 pairs of gloves as simply being gloves of different colors. Picking no matching pair is the same as picking no 2 gloves of the same color. To compute the probability of doing so, we can compute the number of ways to select 8 gloves from different colors, and divide that by the total number of ways to select 8 random gloves out of the 20 gloves.

To compute the number of ways in which we can select 8 gloves from different colors, we can think of the choosing procedure as follows:

1st step- We choose from which 8 colors are we going to pick gloves from. So we have to pick 8 out of 10 colors. This can be done in ${10 \choose 8}$ ways.

2nd step - We now have to choose which glove are we going to pick from each of the chosen colors. Either the left one or the right one. For the first chosen color we have 2 choices, for the second chosen color we have 2 choices, for the third chosen color we have 2 choices, and so on. Therefore the number of ways in which we could choose gloves from the chosen colors is $2^8$

And so the total number of ways in which we could choose 8 gloves from different colors is

${10 \choose 8 }2^8$

Now, the total numer of ways in which we could choose 8 gloves out of the 20 gloves is simply ${20 \choose 8}$

So the probability of picking no mathing pair is

$\frac{{10 \choose 8}2^8}{{20 \choose 8}}\approx 0.091$