Problem 49 - A First Course in Probability

Question

A group of 6 men and 6 women is randomly divided into 2 groups of size 6 each. What is the probability that both groups will have the same number of men?

Step-by-Step Solution

Answer

The probability of forming two groups with an equal number of men in each group is \(\frac{100}{231}\), which is approximately 0.433.

Step 1: Calculate total number of ways to form two groups

Let's start by counting the total number of ways to form two equal-sized groups from the original group of 12 people. We don't care about the gender distribution at this point; we just want to know how many different ways we can divide the group. To do this, we can use the combination formula, which is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where n is the total number of items (people in this case), k is the number of items we want to choose from the total number, and ! is the factorial. In our case, n = 12 (total number of people), and k = 6 (since we want to create a group of 6). \[ C(12, 6) = \frac{12!}{6! (12-6)!} = \frac{12!}{6! 6!} \] Calculating this combination, we get: \[ C(12, 6) = 924 \] So, there are a total of 924 ways to divide the group into two equal-sized groups.

Step 2: Calculate the number of ways to form two groups with equal number of men

Now, we want to find the number of ways to divide the group into two groups of 6 with an equal number of men in each. Since there are 6 men, an equal distribution means we want to have 3 men in each group. To calculate this, we will use combinations again but this time for selecting the men. \[ C(6, 3) = \frac{6!}{3! (6-3)!} = \frac{6!}{3! 3!} \] Calculating this combination, we get: \[ C(6, 3) = 20 \] So, there are 20 ways to select 3 men to be in one of the groups. However, we also have to allocate the remaining men and women. Since there are 3 men and 3 women left, we need to calculate the number of ways to choose 3 women from the 6 women. This will be calculated as: \[ C(6, 3) = \frac{6!}{3! (6-3)!} = \frac{6!}{3! 3!} = 20 \] Now, we can multiply these two combinations because we are selecting 3 men and 3 women, so we use the fundamental counting principle: \[ 20 (ways\, to\, select\, 3\, men) \times 20 (ways\, to\, select\, 3\, women) = 400 \] So, there are 400 ways to divide the group into two equal-sized groups with an equal number of men in each group.

Step 3: Calculate the probability of forming two groups with an equal number of men

To find the probability, we will divide the number of desired outcomes (equal men in both groups) by the total possible outcomes: \[ P(\mathrm{Equal\, number\, of\, men\, in\, both\, groups}) = \frac{400 (\mathrm{desired\, outcomes})}{924 (\mathrm{total\, outcomes})} \] Calculate the probability as a fraction: \[ P(\mathrm{Equal\, number\, of\, men\, in\, both\, groups}) = \frac{400}{924} \] To simplify this fraction, we can divide both numerator and denominator by their greatest common divisor (gcd): GCD(400, 924) = 4. \[ P(\mathrm{Equal\, number\, of\, men\, in\, both\, groups}) = \frac{400/4}{924/4} = \frac{100}{231} \] So, the probability of forming two groups with an equal number of men in each group is 100/231, which is approximately 0.433.