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The U.S. Senate consists of $100$ senators, with $2$ from each of the $50$ states. There are $50$ Democrats in the Senate. A committee of size $10$ is formed, by picking a random set of senators such that all sets of size $10$ are equally likely.

a) Find the expected number of Democrats on the committee.

b) Find the expected number of states represented on the committee (by at least one senator).

c) Find the expected number of states such that both of the state’s senators are on the committee.

  • For part a), I defined $D$ as a random variable for democrats, got the Hypergeometric Distribution to be $D$ ~ $(100, c, d),$ and the expectation to be $E(D) = c(\frac{d}{100}) = \frac{cd}{100}$. Then I plugged in $10$ for $c$ and $50$ for $d$ and got $E(D) = 5.$
  • For part b), I defined $I_j$ as the random variable for the j-th state represented/ not represented in the committee, got $P(I_j = 1) = 1 - P(I_j = 0) = 1 - \frac{\binom{98}{c}}{\binom{100}{c}} = 1 - \frac{(100 - c)(99 - c)}{(100)(99)}$, and finally $E(\sum_{j} I_j) = \sum_{j}E(I_j) = \sum_{j}P(I_j = 1) = 50(1 - \frac{(100 - c)(99 - c)}{(100)(99)})$.

Then I plugged in $10$ for $c$ and got $P(I_j = 1) = 1 - \frac{8010}{9900}$ = $\frac{21}{110}$ and $E(\sum_{j} I_j) = 50(1 - \frac{(100 - 10)(99 - 10)}{(100)(99)})$ = $50(\frac{21}{110})$ = $\frac{105}{11} $

  • For part c), I defined $K_j$ as the random variable for both being from/ not from j-th state in the committee, got $P(K_j = 1) = \frac{\binom{98}{c}}{\binom{100}{c}} = \frac{(100 - c)(99 - c)}{(100)(99)}$ and finally $E(\sum_{j} K_j) = \sum_{j}E(K_j) = \sum_{j}P(K_j = 1) = 50( \frac{(100 - c)(99 - c)}{(100)(99)})$

Then I plugged in $10$ for $c$ and got $P(K_j = 1) = \frac{8010}{9900}$ = $\frac{89}{110}$ and $E(\sum_{j} K_j) = 50(\frac{(100 - 10)(99 - 10)}{(100)(99)})$ = $50(\frac{89}{110})$ = $\frac{445}{11} $.

  • However, I am not sure if my answers are correct. Any help would be much appreciated!
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    $\begingroup$ If I am interpreting your work correctly, you are saying that it is to be expected that there will be (on average) $\frac{445}{11}$ states such that both of the state's senators are on the committee. How can this be when the committee only has 10 positions. Isn't the maximum # of states that could have both of its senators on the committee $10 \div 2 = 5.$ $\endgroup$ Nov 8, 2020 at 7:32
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    $\begingroup$ a and b looks correct. Can you explain the definition of Kj again? $\endgroup$ Nov 8, 2020 at 7:35
  • $\begingroup$ @VishaalSudarsan, yes it is whether both are from the jth state in the committee or not. I am not sure of a different way to explain it sorry. $\endgroup$
    – user102156
    Nov 8, 2020 at 7:43
  • $\begingroup$ @user2661923 I do not know, I would expand on my thought process but now I am confused on part C. $\endgroup$
    – user102156
    Nov 8, 2020 at 7:45
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    $\begingroup$ For a specific senator, that probability is $\frac{1}{10}$. The probability of the next senator from the same state to be on it is $\frac{1}{11}$ and so it is $\frac{1}{110}$. $\endgroup$
    – Math Lover
    Nov 8, 2020 at 7:57

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A) and B) looks fine, for C) I think the right definition for $K_j$ would be let $K_j$ be the indicator variable denoting if there is representation for state $j$ in the committee from both the senators.

Then $P[K_j = 1 ]$ = $\frac{{2 \choose {2}}*{{98}\choose{8}}}{{100}\choose{10}}$

Therefore the expected number of such states in the committee is $50*P[K_j = 1] = \frac{45}{99}$

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