Probability (Selecting at most one of the 3 boys who do not want to serve together to form the committee)

combinatoricsprobability

A class contains 8 boys and 9 girls. The teacher selects 7 of the children at random to form a committee. Each child has equal chance to be selected. If 3 of the boys do not want to serve together, what is the probability that at most one of them is selected to form the committee?

I've approached this question by doing 8𝐶5 + 9𝐶2 / 17𝐶7 as three of the boys do not want to serve with each other but I'm not too sure if I am right.

I would greatly appreciate any help that can point me in the right direction for this question.

Best Answer

Most probably you wanted to write "(8𝐶5 + 9𝐶2) / 17𝐶7".

Note that generally $a+b/c \neq (a+b)/c$. Entering the left expression to calculate the right one into a calculator respecting the order of operations will permanently lead to erroneous results.

Besides this, (8𝐶5 + 9𝐶2) would mean that exactly 5 boys without the 3 problematic ones and 2 girls would form a committee.

A possible approach could be (I use $\binom{n}{k}$ instead of nCk):

  • Number of committees containing none of the $\color{blue}{3}$ problematic boys: $\binom{8\color{blue}{-3}+9}{7} =\binom{14}{7}$
  • Number of committees containing exactly $\color{blue}{1}$ of the problematic boys (choose $1$ of them and choose 6 from the other people): $\binom{3}{\color{blue}{1}}\binom{8\color{blue}{-3}+9}{6} =3\binom{14}{6}$

So, you get $$\frac{\binom{14}{7} + 3\binom{14}{6}}{\binom{17}{7}}$$

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