The probability that exactly one pair of twins exists in 5 people chosen at random from a group of 6 pairs of twins

combinationscombinatoricsprobability

A pyschology experiment involves 6 pairs of twins. In one test, 5 persons are randomly chosen from them. What is the probability that, among the 5 persons, there is exactly one pair of twins?
(a) 2/11
(b) 20/99
(c) 1/5
(d) 3/7
(e) None of the above

My attempt to solve this on my paper was largely unsuccessful, but my simple program in python points towards 20/33 as the answer. As we increase total_iterations, the number approaches 0.606060… which is 20/33.

I have no good approach to reach 20/33 as a solution. I would appreciate if you provide a detailed solution for this problem.
If the answer ends up to not be 20/33, I would appreciate if you can pin poin the problem in my code.
Thank you!

My incorrect solution:
First we choose a pair of twins(A, a), then we choose 2 people(B, C), one each from a different pair, then we choose 1 person form the remaining 8 such that they do not form a pair with either B or C.(D, d, E, e, F, f)

So what I did was:

  1. We can choose any pair
    Combinations for (A, a): 6
  2. We can choose any combination except the 5 pairs
    Combinations of (B, C): 12C5 – 5 = 40
  3. We can choose any 1 of the remaing 8, except two(b, c)
    Combinations for the remaining person = 8 – 2 = 6

Total favourable combinations = 6 * 40 * 6 = 1440
Total combinations = 12C5 = 792

Our final probability is 1440/792 = 1.8181…
This breaks the fundamentals of probability…

Best Answer

As you found, there are $\binom{12}{5}$ ways to select five of the $12$ people.

For the favorable cases, there are six ways to select the pair of twins from which both twins are selected, $\binom{5}{3}$ ways to select from which three of the other five pairs of twins exactly one of the twins will be selected, and two choices for which twin will be selected from each of those three pairs. Hence, the number of favorable cases is $$\binom{6}{1}\binom{5}{3}2^3$$

Therefore, the probability that exactly one pair of twins will be selected when five people are chosen from six pairs of twins is $$\frac{\dbinom{6}{1}\dbinom{5}{3}2^3}{\dbinom{12}{5}} = \frac{20}{33}$$ as your program correctly predicted.

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