Question 697079: Not so much a homework problem as an actual thing I'm grappling with in a gaming group. Your help would be much appreciated!
Four ten-sided dice are rolled. Each die that shows a 6 or lower is a "success", so the chance of a success is 60%, and the chance of a failure is 40%.
The chance of four successes is 0.6^4, or .1296 (12.96%), and the chance of four failures is 0.4^4, or .0256 (2.56%).
What are the chances of exactly one success, exactly two successes, and exactly three successes?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Four ten-sided dice are rolled. Each die that shows a 6 or lower is a "success", so the chance of a success is 60%, and the chance of a failure is 40%.
The chance of four successes is 0.6^4, or .1296 (12.96%), and the chance of four failures is 0.4^4, or .0256 (2.56%).
What are the chances of exactly one success, exactly two successes, and exactly three successes?
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Binomial Problem with n = 4 ; p(1<= x <=6) = 0.6
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P(1 success) = 4C1(0.6)(0.4)^3 = 4*0.6*0.064 = 0.1536
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P(2 successes) 4C2(0.6)^2(0.4)^2 = 6*0.36*0.16 = 0.3456
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P(3 successes) = 4C3(0.6)^3(0.4) = 4*0.216*0.4 = 0.3456
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Cheers,
Stan H.
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