math destroyer question probability

[spoog74]

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if you flip a fair die 3 times what is the probability of obtaining exactly 2 fours?

Im confused as to the explanation of the problem... Destroyer says "to get exactly 3 fours ther are 3 possibilities 44X , 4X4, X44. Thus P = 1/3 * 1/3 * 5/6 = 5/72. My question is where the hell does this 5/6 come from? And why do we need that? Wouldnt it be 1/3*1/3*1/3 which would give you our answer?

Smeone please explain if possible/

 

[carlfusco]

The 5/6 accounts for the role during which you do NOT want to roll a 4... 5/6 of the numbers are NOT a four.

So it works out like this:

Probability of rolling a 4 - (1/6)
Probability of not rolling a 4 - (5/6)

So, like the book said, there are 3 combinations: X44, 4X4, 44X

(5/6*1/6*1/6) + (1/6*5/6*1/6) + (1/6*1/6*5/6)
OR
3(1/6*1/6*5/6)

equals

15/216, reduced to 5/72

I'm not sure where they get the two 1/3s from, but this is a sufficient solution to the problem.