#1**+2 **

wow, I totally messed this one up... have to stop answering these when I'm drinking!

Let's start again shall we

\(P[k] = \dfrac{\dbinom{20}{k}\dbinom{10}{3-k}}{\dbinom{30}{3}}\\ P[(0,1,2,3)] = \large \left(\frac{6}{203},\frac{45}{203},\frac{95}{203},\frac{57}{203}\right)\)

\(E[X] = 0\cdot \dfrac{6}{203} + 1 \cdot \dfrac{45}{203}+2\cdot \dfrac{95}{203}+3\cdot \dfrac{57}{203} =2\)

.Rom Apr 19, 2019

#2**+2 **

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???

CPhill Apr 19, 2019

#4**+1 **

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 **k**iss begins with (k), not every **k**iss 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

GingerAle Apr 22, 2019