Solved – Probability of selecting exactly 2 members of a group of 7, out of 35 people, if 3 people are picked

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There are 7 friends A, B, C, D, E, F, and G that belong to a classroom of 35 students. Three students are chosen from the 35.

What the probability that exactly two of the group of friends is chosen?

Probability that exactly none of the seven friends are chosen?

. . . . .

So I know the total number of combinations of 3 students chosen out of 35 is (35 c 3). That is the denominator for these questions. I can't figure out what the numerator should be.

Is it (7 c 2)/(35 c 3) and (7 c 0)/(35 c 3) ? This does not seem correct.

The number (7 c 2) I chose because I need 2 member from this group of seven, and there are (7 c 2) combinations of getting (AB, BC, AD, FE, etc). On second thought, I might also want to include the possibility ways of getting a single non-friend member for the last slot, so it the numerator:

(7 c 2)*(35 – 7) ?

Then for no friends, it would be (7 c 0)*(35)(34)(33) as the numerator.
Because there are three slots for non-friends?

Best Answer

I am going to take an approach similar to others, but I will focus on deriving the general case and then moving to the particular. Consider instead that you have $N$ people of which $n$ are friends and you pick an arbitrary number $i$ people. You might then ask what is the probability that $j$ of those $i$ people are from the group of friends?

The number of possible states can be found by looking at $N$ from which you choose $i$. This should be straightforward, but in case you need the answer:

$\binom{N}{i}$

We can then calculate the number of states where this is fulfilled. First out of the $n$ friends we pick $j$. Out of the non-friends we pick $i-j$. How many states is this?

$\binom{n}{j} \binom{N-n}{i-j}$

Overall then what is the probability?

$\frac{\binom{n}{j} \binom{N-n}{i-j}}{\binom{N}{i}}$

We can check that our expressions work by summing over $j$ to ensure that the probability sums to 1.

$\sum_j \binom{n}{j} \binom{N-n}{i-j} = \binom{N}{i}$

I will leave it to you to plug in the numbers.

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