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There are 3 math clubs in the school district, with 5, 7, and 8 students respectively. Each club has two co-presidents. If I randomly select a club, and then randomly select four members of that club to give a copy of Introduction to Counting and Probability, what is the probability that two of the people who receive books are co-presidents?

 Aug 6, 2022
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With the 5 student club, there are \({5 \choose 4} = 5\) ways to choose the 4 students. Of these, there are \({2 \choose 2} \times {3 \choose 2} = 3\) ways to select both co-presidents. Both co-presidents and be chosen in 1 way, and there \({3 \choose 2} = 3\) ways to choose the remaining students. The probability of selecting this club is \({1 \over 3}\). So, the probability for this club is \({3 \over 5} \times {1 \over 3} = {1 \over 5}\)

 

With the 7 student club, there are \({7 \choose 4} = 35 \) ways to choose the 4 students. Of these, there are \({2 \choose 2} \times {5 \choose 2} = 5\) ways to select both co-presidents. Likewise, the probability of selecting this club is \({1 \over 3}\). So, the probability for this club is \({1 \over 7} \times {1 \over 3} = {1 \over 21}\)

 

With the club with 8 students, there are \({8 \choose 4} = 70\) ways to choose the 4 students. There are also \({2 \choose 2} \times {6 \choose 2} = 15\) ways to choose both co-presidents. 

So, the probability is \({3 \over 14} \times {1 \over 3} = {1 \over 14}\)

 

Adding everything up, we find the total probability is \({1 \over 5} + {1 \over 21} + {1 \over 14} = \color{brown}\boxed{67 \over 210}\)

 Aug 6, 2022
edited by BuilderBoi  Aug 6, 2022

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