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a) Suppose we have a bag with 10 slips of paper in it. Eight slips have 3 on them and the other two have 8 on them. How many 8's do we have to add to make the expected value equal to 6?

 

b) Suppose we have a bag with 10 slips of paper in it. Eight slips have 3 on them and the other two have 8 on them. How many 8's do we have to add before the expected value is at least 7?

 Jun 3, 2023
 #1
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a)  If there are 10 slips of paper in the bag, and 8 of them have 3 on them, then the expected value is 3. If we add one 8 to the bag, then the expected value becomes (8 * 1/11) + (3 * 10/11) = 3.63636..., which is still greater than 6. If we add two 8's to the bag, then the expected value becomes (8 * 2/12) + (3 * 10/12) = 5, which is equal to 6. Therefore, we need to add 2​ 8's to the bag to make the expected value equal to 6.  The answer is 2.

 Jun 3, 2023
 #2
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Thanks for the help! I already tried that answer and it's incorrect.

Guest Jun 3, 2023
 #3
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\(Expected value = \frac{n_3\times3+n_8\times8}{n_3+n_8}\\ a) 6 = \frac{8\times3+(2+n)\times8}{8+2+n}\)

 

Solve for n.

 Jun 3, 2023

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