Probability of Chosen Socks: A Detailed Exploration of Pairs Among 20 Distinct Socks

Imagine a situation where you have a drawer containing 20 distinct pairs of socks. In this scenario, you randomly pick 6 socks. A common question arises: what is the probability that you will pick a specific favorite pair of socks? This problem involves the application of combinatorial mathematics and probability theory. Let’s explore this in more detail.

Understanding the Problem and Initial Assumptions

First, let's state the problem clearly: what is the probability that you will choose your favorite pair of socks when picking 6 socks from a drawer containing 20 distinct pairs of socks? It is essential to note that for the sake of simplicity, we assume that no non-matching pair of socks is selected.

Probability Calculation for the Favorite Pair

The calculation for this problem can be broken down as follows:1. The total number of socks is 40 (20 pairs).2. We are selecting 6 socks without replacement.3. We want to find the probability that the favorite pair is among the 6 socks chosen.

Step-by-Step Solution

To calculate the probability, let's consider the following steps:

Number of ways to choose the favorite pair: Since the favorite pair is a specific pair, there is only 1 way to choose it. Number of ways to choose the remaining 4 socks from the remaining 38 socks: After selecting the favorite pair, we need to pick 4 more socks from the remaining 38 socks (39 pairs with one pair already chosen). The number of ways to do this is given by the combination formula C(n, k) n! / [k!(n-k)!], where n is the total number of items, and k is the number of items to choose. [ C(38, 4) frac{38!}{4!(38-4)!} ] After simplifying, we get: [ C(38, 4) frac{38 times 37 times 36 times 35}{4 times 3 times 2 times 1} ] [ C(38, 4) 73,815 ] Total number of ways to choose 6 socks from 40 socks: This is given by the combination formula for choosing 6 socks from 40 socks. [ C(40, 6) frac{40!}{6!(40-6)!} ] After simplifying, we get: [ C(40, 6) frac{40 times 39 times 38 times 37 times 36 times 35}{6 times 5 times 4 times 3 times 2 times 1} ] [ C(40, 6) 383,838,000 ]

The probability P(favorite pair is chosen) is the ratio of the number of favorable outcomes to the total number of possible outcomes.

[ P(text{favorite pair chosen}) frac{C(38, 4)}{C(40, 6)} ]

Substituting the values:

[ P(text{favorite pair chosen}) frac{73,815}{383,838,000} approx 0.0001924 ]

This means the probability of picking the favorite pair when randomly selecting 6 socks from a drawer of 20 distinct pairs is approximately 0.01924%.

Probability of No Complete Pair

Next, let's explore a related question: what is the probability that you will get no complete pair of socks when picking 6 socks from the drawer containing 20 distinct pairs?

Step-by-Step Solution

Number of ways to choose 6 socks such that no pair is formed: This is a classic problem of forming a combination where no two socks from the same pair are chosen. The number of ways to choose 6 socks from 20 pairs such that no two socks form a pair is given by the following: [ binom{20}{6} times 2^6 ] Here, (binom{20}{6}) is the number of ways to choose 6 different pairs out of 20, and (2^6) is the number of ways to choose one sock from each of the 6 pairs (2 choices per pair, 6 pairs). [ binom{20}{6} frac{20!}{6!(20-6)!} ] After simplifying, we get: [ binom{20}{6} 38,760 ] So, [ binom{20}{6} times 2^6 38,760 times 64 2,488,960 ] Total number of ways to choose 6 socks from 40 socks: This is the same as before, which is (C(40, 6) 383,838,000).

The probability P(no complete pair is chosen) is the ratio of the number of favorable outcomes to the total number of possible outcomes.

[ P(text{no complete pair chosen}) frac{2,488,960}{383,838,000} approx 0.00647 ]

This means the probability of picking 6 socks such that no complete pair is formed is approximately 0.647%.

Conclusion

In summary, the probability of picking a specific favorite pair of socks when randomly selecting 6 socks from a drawer of 20 distinct pairs is approximately 0.01924%. Meanwhile, the probability of picking 6 socks such that no complete pair is formed is approximately 0.647%. These calculations are essential for understanding the combinatorial nature of probability in real-life scenarios.