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avatar+554 

Suppose I have a bag with 10 slips of paper in it. Eight of these have a 2 on them and the other two have a 4 on them.

How many 4's do I have to add before the expected value is at least 3.5?

 Nov 24, 2018
 #1
avatar+3895 
+2

\(E[X] = 2 \cdot \dfrac{8}{n}+4\cdot \dfrac{n-8}{n} = \\ \dfrac{16}{n}-\dfrac{32}{n} + 4 = \\ 4 - \dfrac{16}{n}\)

 

\(3.5 = 4 - \dfrac {16}{n}\\ \dfrac{16}{n} = \dfrac 1 2\\ n = 32\\ \text{Thus we must add }32-10 = 22 \\ \text{4's to the bag}\)

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 Nov 24, 2018
 #2
avatar+95884 
+1

Call the number of additional slips showing a '4", x

 

So we have

 

2 [  8 / (10 + x) ]   +  4  [ 2 + x ] / [ 10 + x ]   = 3.5         simplify

 

16 / (10 + x)   +  [ 8 + 4x] / (10 + x)   = 3.5

 

16  +  8  +  4x    =   3.5 ( 10 + x)

 

24 + 4x    =  35 + 3.5x

 

.5x   =   11

 

x =  22

 

 

cool cool cool

 Nov 25, 2018

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