SOLUTION: The probability of rolling exactly three 4's in 5 rolls of a fair die

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Question 450603: The probability of rolling exactly three 4's in 5 rolls of a fair die
Found 3 solutions by ewatrrr, stanbon, edjones:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
The probability of rolling exactly three 4's in 5 rolls of a fair die
P(rolling a four)= 1/6
P(not rolling a four) = 5/6
1/6*1/6*1/6*5/6*5/6
distinguishable orders of the five rolls/3fours and '2' others:5!/[3!*2!]=10
P(exactly 3 fours in five rolls) = 10*(1/6*1/6*1/6*5/6*5/6)
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
The probability of rolling exactly three 4's in 5 rolls of a fair die
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Binomial Problem with n = 5 and p = 1/6
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P(x = 3) = 5C3(1/6)^3(5/6)^2 = 10*(25/6^5) = 0.0322
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Cheers,
Stan H.
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Answer by edjones(8007)   (Show Source): You can put this solution on YOUR website!
p=probability of rolling a four
p=1/6, q=5/6
(p+q)^5 expand
5C3=10 coefficient
10p^3q^2=.03215 third term of the expansion.
.
Ed

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