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Chapter 2: Problem 49

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?

Short Answer

Expert verified
The probability that both groups will have the same number of men when 6 men and 6 women are randomly divided into 2 groups of size 6 each is \(\frac{5}{66}\).

Step by step solution

01

Calculate the total number of ways to divide the 12 people into 2 groups of size 6.

To choose a group of 6 people out of the total 12 people, we can use combination. Combination formula is: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where n is the total number of elements and k is the number of elements we want to choose. In our case, n = 12 (total number of people) and k = 6 (size of each group): \(\binom{12}{6} = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!}\)
02

Calculate the number of ways to divide the people with equal number of men and women in each group.

Both groups will have 3 men and 3 women. We can then focus on selecting 3 men out of the 6 men and do the same for the women. First, we calculate the number of ways to select 3 men out of 6: \(\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}\) Similarly, we calculate the number of ways to select 3 women out of 6: \(\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}\) To obtain the total arrangements with an equal number of men and women in each group, we multiply the number of ways to select men and women: \((\frac{6!}{3!3!})(\frac{6!}{3!3!})\)
03

Calculate the probability of having equal number of men and women in each group.

To find the probability, we divide the number of favorable outcomes (equal number of men and women in each group) by the total number of possible outcomes (all possible ways to divide the 12 people into 2 groups of size 6). Probability = \(\frac{(\frac{6!}{3!3!})(\frac{6!}{3!3!})}{\frac{12!}{6!6!}}\) Now, we can cancel the common terms and simplify: Probability = \(\frac{6! * 6! * 6! * 6!}{3! * 3! * 3! * 3! * 12!}\)
04

Final answer.

After further simplification, we get: Probability = \(\frac{5}{66}\) Therefore, the probability that both groups will have the same number of men when 6 men and 6 women are randomly divided into 2 groups of size 6 each is \(\frac{5}{66}\).

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Most popular questions from this chapter

Consider the matching problem, Example \(5 \mathrm{~m}\), and define \(A_{N}\) to be the number of ways in which the \(N\) men can select their hats so that no man selects his own. Argue that $$ A_{N}=(N-1)\left(A_{N-1}+A_{N-2}\right) $$ This formula, along with the boundary conditions \(A_{1}=0, A_{2}=1\), can then be solved for \(A_{N}\), and the desired probability of no matches would be \(A_{N} / N !\) HINT: After the first man selects a hat that is not his own, there remain \(N-1\) men to select among a set of \(N-1\) hats that does not contain the hat of one of these men. Thus there is one extra man and one extra hat. Argue that we can get no matches cither with the extra man selecting the extra hat or with the extra man not selecting the extra hat.

Prove the following relations. $$ F=F E \cup F E^{c}, \text { and } E \cup F=E \cup E^{c} F $$

A basketball team consists of 6 frontcourt and 4 backcourt players. If players are divided into roommates at random, what is the probability that there will be exactly two roommate pairs made up of a backcourt and a frontcourt player?

A town contains 4 people that repair televisions. If 4 sets break down, what is the probability that exactly \(i\) of the repairers are called? Solve the problem for \(i=1,2,3,4\). What assumptions are you making?

An ordinary deck of 52 cards is shuffied. What is the probability that the top four cards have (a) different denominations; (b) different suits?
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