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- anonymous

A diamond merchant has a chance to buy one of 4 diamonds and wants to sell it immediately. One costs only $1000, but he has a 90% chance of getting $5000 for it (or not selling it at all). The second costs $10,000, and he has a 60% change of selling it for $25,000 (or not selling it at all). The third costs $20,000, but he has a 50% chance of selling it for $60,000 (or not selling it at all). The last diamond costs $50,000, but he has a 1% chance of selling it for $1,000,000 (or not selling it at all). Which diamond gives him the best expected value?

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- anonymous

- schrodinger

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- anonymous

options are :
the one for $1000
the one for $10,000
the one for $20,000
the one for $50,000

- kropot72

First convert the percentages to decimals, for example 90% converts to a probability of 0.9.
Then multiply the profit by the probability to get the expected value.
Cost Profit Probability Expected Value
$1000 5000 - 1000 = $4000 0.9 $4000 * 0.9 = $3600
Now you can complete the table to find the correct choice of answer.

- kropot72

@shavu1 Are you there?

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- anonymous

yes

- anonymous

wait what chart

- anonymous

- kropot72

I have put the details for the first diamond in the above table to show you the method. Now you need to put in the details for the other three diamonds.

- anonymous

am i looking for what expected value gives me 1mil

- anonymous

1.3600
2.9000
3.20000
4.95000

- anonymous

i calculated and this what i got for the last 3

- kropot72

|dw:1442257433971:dw|

- anonymous

i finished the chart

- anonymous

wouldnt the right answer be D it has the highest expected value

- kropot72

What was your calculated expected value for D?

- anonymous

95,000

- kropot72

Not correct. Check your calculation again.

- anonymous

shouldnt it be 1,000,000-50,000= 95,000 x .01

- anonymous

nvm D is 950

- anonymous

the best answer would be C since its 20,000

- kropot72

The calculation for D is : $950,000 * 0.01 = $9500

- anonymous

so its C

- kropot72

Yes, C gives the highest expected value.

- anonymous

THANKS :d

- kropot72

You're welcome :)

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