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Suppose we have a bag with 10 slips of paper in it. Eight slips have a 3 on them and the other two have an 8 on them.

 

a) What is the expected value of the number shown when we draw a single slip of paper?

b) What is the expected value of the number shown if we add one additional 8 to the bag?

c) What is the expected value of the number shown if we add two additional 8's (instead of just one) to the bag?

d) How many 8's do we have to add to make the expected value equal to 6?

e) How many 8's do we have to add before the expected value is at least 7?

 May 13, 2024
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a) \(\dfrac{8 \cdot 3 + 2 \cdot 8}{10} = \dfrac{24 + 16}{10} = \dfrac{40}{10} = \boxed{4}\)

b) \(\dfrac{8 \cdot 3 + 3 \cdot 8}{11} = \dfrac{24 + 24}{11} = \boxed{\dfrac{48}{11}}\)

c) \(\dfrac{8 \cdot 3 + 4 \cdot 8}{12} = \dfrac{24 + 32}{12} = \dfrac{56}{44} = \boxed{\dfrac{14}{11}}\)

d) \(\begin{align*} \dfrac{8 \cdot 3 + (2 + x) \cdot 8}{10 + x} &= 6 \\ \dfrac{24 + 16 + 8x}{10 + x} &= 6 \\ \dfrac{40 + 8x}{10 + x} &= 6 \\ 40 + 8x &= 60 + 6x \\ 40 + 2x &= 60 \\ 2x &= 20 \\ x &= \boxed{10} \end{align*}\)

e) \(\begin{align*} \dfrac{8 \cdot 3 + (2 + x) \cdot 8}{10 + x} &= 7 \\ \dfrac{24 + 16 + 8x}{10 + x} &= 7 \\ \dfrac{40 + 8x}{10 + x} &= 7 \\ 40 + 8x &= 70 + 7x \\ 40 + x &= 70 \\ x &= \boxed{30} \end{align*}\)

 May 13, 2024

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