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A committee of $5$ members will be chosen from a group of $10$ teachers and 5 students. What is the probability that the committee will have 3 teachers and 2 students? My solution was this :($10C3$ x $5C2$) / $15C5$ . My professor told me that I still need to multiply something to get my answer right which is now my current problem. What do I need to multiply? UPDATE: Ill be meeting my professor later to get the correct answer and I'll share it here.

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    i think ur answer is right – 2017-02-07
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    It is still lacking one factor though by which I need to multiply – 2017-02-07
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    what that is the answer given by ur professor? – 2017-02-07
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    I am thinking that it has something to do with the arrangement? – 2017-02-07
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    that is my solution, my professor didn't tell us the correct answer – 2017-02-07
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    it you consider arrangements, it will equal apply in numerator and denominator and no effect – 2017-02-07

3 Answers 3

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3 teachers from 10 teachers.

$$\binom{10}{3}$$

2 students from 5 students.

$$\binom{5}{2}$$

Committee with 3 teachers and 2 students.

$$\binom{10}{3} \times \binom{5}{2}$$

Total ways to select 5 persons from 15.

$$\binom{15}{5}$$

Probability = $$\frac{\binom{10}{3} \times \binom{5}{2}}{\binom{15}{5}}$$

Same as your answer. So your answer is correct. No need to multiply something.

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Note that if we introduce a descrete random variable $X$ which describes the number of teachers in the committee, then $X$ follows hypergeometric distribution with a probability mass function $P(X=k)=\frac{{{K}\choose{k}}\cdot{{N-K}\choose{n-k}}}{{{N}\choose{n}}}$ for $K$ number of teachers, $N-K$ number of students and $n$ number of people in the committee.

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    so the answer is still the same? – 2017-02-08
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    Why don't you use numbers instead of variables and you will see... – 2017-02-08
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You need to multiply by $1$. (Your answer is already correct)