Problem 52 - A First Course in Probability

Question

A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be (a) no complete pair? (b) exactly 1 complete pair?

Step-by-Step Solution

Answer

The probability of selecting 8 shoes with (a) no complete pair is \(0.045\) and (b) exactly 1 complete pair is approximately \(0.454\).

Step 1: Calculate the number of ways to select 8 shoes without a complete pair

To achieve this, we need to ensure that we select 1 shoe from each of 8 different pairs and none from the remaining 2 pairs. There are 10!/(2!8!) ways to select 8 pairs out of 10 pairs. From each of the 8 pairs, there are 2 choices to pick 1 shoe (as there are 2 shoes per pair), so there are 2^8 possible shoe selections.

Step 2: Calculate the total possible ways of selecting 8 shoes out of 20

Use the binomial coefficient formula (also known as 'n choose k' or combinations) which is: \({n \choose k}= \frac{n!}{k!(n-k)!}\), where n=20 shoes and k = 8 shoes to be selected. It gives us the total number of ways to select 8 shoes from 20 shoes: \({20 \choose 8} = \frac{20!}{8!(20-8)!}\)

Step 3: Compute the probability of having no complete pairs

Divide the number of ways to select 8 shoes without a complete pair by the total possible ways of selecting 8 shoes: \(\frac{ \frac{10!}{2!8!} \times 2^8}{\frac{20!}{8!(20-8)!}}\) After simplifying the math, we get the probability to be \(\frac{45}{990}\) or \(0.045\). (b) Exactly 1 complete pair

Step 1: Calculate the number of ways to select 8 shoes having exactly 1 complete pair

From the 10 pairs, we choose one pair to have a complete pair which gives us 10 possibilities. Then, we select 1 shoe from each of 6 different pairs out of the remaining 9 pairs which gives us 9!/(3!6!) ways. From each chosen pair, we have 2 possibilities of picking a shoe. So there are \(10 \times \frac{9!}{3!6!} \times 2^6\) ways to select the shoes.

Step 2: Compute the probability of having exactly 1 complete pair

Divide the number of ways to select 8 shoes having exactly 1 complete pair by the total possible ways of selecting 8 shoes (from step 2 of part a): \(\frac{10 \times \frac{9!}{3!6!} \times 2^6}{\frac{20!}{8!(20-8)!}}\) Simplify the math to find the probability, which is \(\frac{135}{297}\) or approximately \(0.454\). So, the probability of selecting 8 shoes with (a) no complete pair is 0.045 and (b) exactly 1 complete pair is approximately 0.454.