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# help

0
164
2
+740

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
+5172
+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
+101424
+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

Nov 25, 2018