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Mandarinka [93]
3 years ago
9

two children, moe and joe, are allowed to select candy from a plate of nine pieces of candy. Moe, being younger, is allowed to c

hoose first but can only take two candies. Joe is then allowed to take three of the remaining candies. Joe complains that he has fewer choices than Moe. Is joe correct? How many choices will each child have?
Mathematics
1 answer:
Ugo [173]3 years ago
3 0
Joe is correct because Moe could choose two pieces but had a nine piece variety where as Joe could only choose from 7 pieces of candy
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Answer:

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alukav5142 [94]

Answer: a) -$0.19, b) -$111.72 .

Step-by-step explanation:

Since we have given that

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a) Find the expected value of the proposition.

Expected value of success in next 2 free throws = \dfrac{390}{434}\times \dfrac{391}{435}=0.8077

Expected value would be

0.8077\times 40+(1-0.8077)\times -169\\\\=32.308-32.4987\\\\=-\$0.19

b)  If you played this game 588 times how much would you expect to win or lose?

Number of times they played the game = 588

So, Expected value would be

588\times -0.19\\\\=-\$111.72

Hence, a) -$0.19, b) -$111.72

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