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I'm having trouble understanding the answers to these questions:

Ten students A, B... are in a class. A committee of 3 is chosen from the class. Find the probability that

  1. A belongs to the committee = 3/10

  2. B belongs to the committee = 3/10

  3. A and B belong to committee = 1/15

  4. A or B belong to the committee = 8/15

Where is 3/10 coming from? Why wouldn't it be 1/10? Thank you !

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    How many committees are possible? There are $\binom{10}{3}$ equally likely committees possible. How many committees contain person $A$? $\binom{9}{2}$ contain person $A$. Taking the ratio, we have the probability that $A$ is in the committee is $\binom{9}{2}/\binom{10}{3}$. What is this value simplified? – 2017-02-12

2 Answers 2

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The probability of forming a committee under some condition $X$ is given by

$$ \frac{\text{Number of committees that favor } X}{\text{Total number of committees}} $$

Total number of committees that can be formed from 10 students by selecting 3 students $= \binom{10}{3} = \frac{10}{3}\binom{9}{2} = \frac{10\times 9}{3\times 2}\binom{8}{1}$.

  1. If $A$ must be in the committee, total number of such committees = $\binom{9}{2}$.

  2. Same as the previous case, $\binom{9}{2}$.

  3. If both $A$ and $B$ must be in the committee, then we have freedom only in choosing the other member. Hence, number of such committees $= \binom{8}{1}$.

  4. Total number of such committees is just the sum of the number of favorable committees in the previous cases.

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The members of the class can be represented as a list of 10 items: $$\{\underbrace{\star,\star,\star}_{\text{3 committee members}},\underbrace{\unicode{x1F6B9},\unicode{x1F6B9},\dots,\unicode{x1F6B9}}_{\text{7 non-committee members}}\}.$$ Here $\star$ and $\unicode{x1f6b9}$ stand for committee and non-committee members respectively. Student $A$ can be placed with equal probability into one of the above 10 places. 3 of them represent committee member. That's where the answer $3/10$ from.