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Why is this the wrong approach to solve this problem?

"There are 65 students. 20 of them are sophomores, 20 are freshmen, 15 are juniors and 10 are seniors. When picking a 4 student committee, calculate the probability that at least 1 of the students is a senior."

So I go by this: $$P(E) = \frac{|E|}{|S|}$$. Where |E| is the number of possibilities for the event I want, and |S| is the number of possibilities in the sample space.

So $$|E| = 1 * C(64, 3)$$. I pick the senior first (no choice), then I fill the last 3 spots with all possible combination of the rest of the 64 students.

$$|S| = C(65, 4)$$ i.e. the number of possible committees. Inserting this into the formula for $P(E)$ we get $\frac{4}{65}$. I know this answer is wrong, but I'm wondering why?

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  • $\begingroup$ It is quite a bit easier to first find the probability of no senior. In your calculation, why is there only one way to pick a senior? And if you change that $1$ to $10$, you will be multiple counting committees with more than one senior. $\endgroup$ – André Nicolas Oct 18 '14 at 20:24
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In the first instance, there are ten possible seniors, so you should multiply $E$ by ten.
On the other hand, if there are two seniors, you have counted them once each, so you have overestimated $E$.
It gets convoluted. Start again and count how many ways to have no seniors, then all the other possibilities have at least one senior.

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Define $X$ to be the number of seniors in the $4$ student committee. You want to calculate the probability $$P(X\ge 1)=1-P(X=0)=1-\frac{\binom{55}{4}\cdot\binom{10}{0}}{\binom{65}{4}}=1-0.5037=0.4963$$


In case you are familiar with distributions of random variables, note that $X$ is a hypergeometric random variable with parameters $N=65$, $k=10$ and $n=4$.

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    $\begingroup$ Perfect analysis. Really should be the answer :) $\endgroup$ – David W May 15 '15 at 20:09

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