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March 28, 2015

March 28, 2015

Posted by **Jon** on Thursday, March 6, 2008 at 1:13pm.

- Algebra -
**drwls**, Thursday, March 6, 2008 at 2:05pmProbability all 5 are reduced is

P(5) = (1/4)^5 = 0.000996

Probability that none are reduced is

P(0) = (3/4)^5 = 0.3773

Probability that 3 out of 5 are reduced is

P(3) = (1/4)^3*(3/4)^2* [5!/(2!*3!]= 0.0879

P(1) = (1/4)(3/4)^4*5 = 0.3955

P(2) = (1/4)^2*(3/4)^3*[5!/(2!*3!)] = 0.2637

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