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A bin consists of 4 \$10 bills, 3 \$20 bills and 1 \$100 bill. You select 7 bills out of this bin uniformly at random without replacement. Find the expected value of the sum of all the bills you select.

What would be the best way to approach this question? Can I use linearity of expectation? Does exchangeability have anything to do with this question?

Thanks :)

chae1fan
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    Yes, linearity of expectation is perfectly fine to use here. What is the expected value of the "first" bill you draw? Of the "second"? For a completely different approach, note that there are only 8 bills to begin with and if you select 7 bills without replacement then that is equivalent to selecting only one bill to leave behind and taking the rest. – JMoravitz Nov 29 '23 at 13:53
  • Thanks for your response! I struggle to intuitively understand why linearity of expectation works even when events are not independent. If i'm not mistaken, the answer can be found from finding the expected value of the first draw (which is 25), then multiplying this by 7 to obtain $175. By this logic, does that mean the expected value of all the draws is $25? This does not make sense to me intuitively as subsequent draws are dependent on the outcomes of the prior draws. – chae1fan Nov 29 '23 at 15:24
  • If you draw two cards, what is the probability that the second card is a queen – JMoravitz Nov 29 '23 at 15:25
  • Reworded differently for your specific case... imagine the bills all have different serial numbers on them... Well, which specific serial number pulled is more likely than any other to be pulled in the second draw? None of them, right? If we aren't conditioning on what the first draw is, the second draw is just as likely to be any of the specific bills... same with later draws. Since we are talking about the expected value of the 2nd bill... and are not talking about the conditional expected value of the 2nd bill... we see that it will be identical to the first bill, hence the result. – JMoravitz Nov 29 '23 at 15:30

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The total value of bills is $4\times 10+3\times 20+1\times 100=200$. The value $X$ of the 7 bills you draw is $X=200-Y$ where $Y$ is the value of the last remaining bill. Computing $\mathbb{E} Y$ is easy. Now you can work out $\mathbb{E}X$.

van der Wolf
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