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I thought I knew combinations pretty well, until recently I came across this question:

In a high school debating team consisting of 2 freshmen, 2 sophomores, 2 juniors, and 2 seniors, two students are selected to represent the school at the state debating championship. The rules stipulate that the representatives must be from different grades, but otherwise the 2 representatives are to be chosen by lottery. What is the probability that the students selected will consist one freshman and one sophomore?

The way I approached this problem was to compute $\frac{N}{D}$, where

  • $N =$ number of ways a sophomore and freshman can be chosen
  • $D$ = total number of ways students can be chosen from different grades

The I broke it down as follows:

  • $D$ = total number of ways to choose 2 students - number of ways to choose students from the same grade. So, $D = C(n=8, k=2) - 4 = 28 - 4= 24$
  • $N$ = 1, because there is only one way to choose a sophomore and a freshman.

But this gets me $\frac{1}{24}$, which is incorrect. I understand there are other ways of doing this problem, but I'm interested in knowing where I went wrong in my logic/approach of the problem?

Thanks.

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You have correctly calculated the denominator. However, there are four ways to select a freshman and a sophomore since there are two ways to select a freshman and two ways to select a sophomore. Hence, the probability that a freshman and a sophomore are selected for the competition is $$\Pr(\text{selected students are a freshman and a sophomore}) = \frac{\binom{2}{1}\binom{2}{1}}{\binom{8}{2} - \binom{4}{1}\binom{2}{2}} = \frac{4}{24} = \frac{1}{6}$$ assuming each individual is equally likely to be chosen.

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    $\begingroup$ Right...I realized this myself too. In fact, there's an easier way to compute the denominator if you consider multiplying these two together: 1) number of ways to choose which two different grades to choose from = $C(4,2) = 6$, and 2) number of ways to choose the 2 individuals from those 2 grades = 4 (as you pointed out). 4*6 = 24. thanks $\endgroup$ – Sother Aug 26 at 15:53
  • $\begingroup$ @Sother Good observation. $\endgroup$ – N. F. Taussig Aug 26 at 18:59

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