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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?

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    $\begingroup$ Please add the self-study tag, and read its tag wiki, modifying your question as needed; specifially, identify why you chose those numerators and why they then don't seem correct to you. With those changes you should be able to get some hints and guidance. $\endgroup$ – Glen_b Apr 23 '15 at 2:12
  • $\begingroup$ You may also find it useful to read about the hypergeometric distribution $\endgroup$ – Glen_b Apr 23 '15 at 2:14
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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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Here is the logic of the probabilities. The $n \choose k$ representation is hidden because I believe that is what you are going for. Try to work it out

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

2 of the friends chose means $$P(Friend~draw1) = \frac{7}{35}$$ and then $$P(Friend~draw2) = \frac{6}{34}$$ and finally $$P(Not~ Friend~draw3) = \frac{28}{33}$$ This give a probability of $$P(2~friends~on~3~draws) = \frac{7}{35}\times\frac{6}{34}\times\frac{28}{33}$$ However you can draw {friend,friend,Not friend} or {friend,Not friend,friend} or {Not friend,friend,friend} there are $3 \choose 2$ ways of picking 2 friends out of 3. In general we can write the probability of picking from a group as $\frac{f}{f+n}$ and $\frac{n}{f+n}$ where $f$ and $n$ are the number of remaining people in each group respectively. $$~~P(friend,friend,Not friend) = \frac{7}{35}\times\frac{6}{34}\times\frac{28}{33}$$ $$~~P(friend,Not friend,friend) = \frac{7}{35}\times\frac{28}{34}\times\frac{6}{33}$$ $$+~P(Not friend,friend,friend) = \frac{28}{35}\times\frac{7}{34}\times\frac{6}{33}$$ Since multiplication is both associative and commutative these 3 probabilities are equivalent. Therefore we can write $$~~P(2 ~friends ~ on ~3 ~draws) = \frac{7}{35}\times\frac{6}{34}\times\frac{28}{33}\times 3$$

And this can be expressed in chooses as $$\frac{{7 \choose 2}{28\choose1}}{35 \choose 3}$$

2) Probability that exactly none of the seven friends are chosen?

For the first draw you have $$P(Not~ Friend~draw1) = \frac{28}{35}$$ on the second draw you now have 27 non-friends and 34 remaining students $$P(Not~ Friend~draw2) = \frac{27}{34}$$ and on the third draw you have again one less friend and one less student $$P(Not~ Friend~draw3) = \frac{26}{33}$$ So the probability of drawing 3 non-friend is $$P(3~ non~ friends~ out ~of~ 3) = \frac{28}{35}\times\frac{27}{34}\times\frac{26}{33}$$

This is the same as doing $$\frac{28 \choose 3}{35 \choose 3}$$

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    $\begingroup$ The rationale for (1) appears to be incomplete, because when the order in which friends and non-friends are encountered is changed, the probabilities change too. For instance, the chances involved when the sequences is non-friend, friend, friend are $28/35$, $7/34$, and $6/33$ respectively. Individually, they differ from the chances you used. Thus, you also need to demonstrate that despite these differences, their product remains the same. $\endgroup$ – whuber Jun 8 '16 at 17:43
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    $\begingroup$ @whuber Yeah I kind of skipped a mental step. Hope the edit clears up the logic $\endgroup$ – Marsenau Jun 8 '16 at 18:05
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Encode the problem: you have an urn with 35 white balls. You remove 7 balls, paint them blue, and put them back in the urn. You mix all the balls in the urn. If you draw 3 balls at random from the urn without replacement, what is the probability of getting exactly 2 blue balls? ($\approx 9\%$) The hypergeometric distribution is your friend here. The second question is similar: what is the probability of getting 3 white balls? ($\approx 50\%$)

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7 friends to shose from. What the probability that exactly two of the group of friends is chosen? Ans.(7C2) 1 person left to choose other than from those friends. Total number of students are 35. Two are already selected so from 35 32 remains. now 1 student more is needed from these 32. (1C32). Now it would be : (7C2)*(1C32).

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    $\begingroup$ Are 2 removed from the 35? It seems like you would remove the 7 friends. This is in agreement with vandermonde identity. To be a probability it you would need to divide by 35 c 3. $\endgroup$ – Kitter Catter Jun 8 '16 at 17:22

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