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January 26, 2015

January 26, 2015

Posted by **Rosmery** on Tuesday, September 18, 2012 at 4:25pm.

- Probability and statistics -
**Reiny**, Tuesday, September 18, 2012 at 5:29pmProb(2) = 1/36

prob(3) = 2/36

prob(4) = 3/36

prob(12) = 1/36

prob(11) = 2/36

prob(10) = 3/36

total of above , your cases of winning = 12/36

so the prob of remaining cases = 24/36

expected value of game

= (12/36)(5) + (24/36)(-5)

= (1/3)(5) - (2/3)(5

= -5/3

You would be expected to lose $1.67

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