+6251 wow, I totally messed this one up... have to stop answering these when I'm drinking! 
Let's start again shall we
\(P[k] = \dfrac{\dbinom{20}{k}\dbinom{10}{3-k}}{\dbinom{30}{3}}\\ P[(0,1,2,3)] = \large \left(\frac{6}{203},\frac{45}{203},\frac{95}{203},\frac{57}{203}\right)\)
\(E[X] = 0\cdot \dfrac{6}{203} + 1 \cdot \dfrac{45}{203}+2\cdot \dfrac{95}{203}+3\cdot \dfrac{57}{203} =2\)
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+129852 One question,Rom...[I'm not too familiar with this....]
Don't we have
(0)(57/203) + 1(95/203) + 2(45/203) + 3(6/203) by the summation formula???
+2489 Rom, the RHS your equation was correct; it’s the LHS (formula) that has the error.
\(E[X] = \underbrace {\sum \limits_{k=0}^3 k \cdot P[k]}_{LHS}= \frac{57}{203} + 2 \cdot \frac{95}{203} + 3 \cdot \frac{45}{203} + 4 \cdot \frac{6}{203} = 2\\ \)
The general formula for expectation:
\( {E} [X]=\sum \limits _{i=1}^{k}x_{i} \cdot p_{i} ,|, \small \text{Where (k) is the total number of events, and (i) is its index, multiplied by its probability.} \)
You (intuitively) corrected for this on the RHS. I assume the different use of the (k) variable derailed your intent. In the moment, we intellectually know that while every kiss
begins with (k), not every kiss is the same. But... it’s easier to analyze that after the fact, though. 
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For this question, it’s easy to verify that two (2) is the expected number of juniors in the committee: The 20 juniors represent (2/3) of the total sample space of 30. (2/3) of (a random sample of) three (3) = 2; the expected number of juniors.
GA
