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Simplify this further for the lack of A/C's.. = (5^4 + 4*5^3)/6^4
= (5*5^3 + 4*5^3)/6^4
= (5+4)*5^3/6^4
= 9*5^3//(3*2)^4
= 3^2*5^3/3^4*2^4
= 5^3/3^2*2^4
= 125/9*16
= 125/144
very high probability indeed

Yes I am reviewing probability from a college text book. I am trying random problems in the back - some may not be direct gmat material but its practice nonetheless.

bewakoof,
The probability of getting at least one 3 in three rolls is the same as:

1 - the probability of getting no 3's in 3 rolls...

Since there is a 1/6 probability that you will get at least one 3, the probability that you will get no 3s is 5/6. Thus, the prob that on all 3 rolls the die will not yield a 3 is:

5/6 * 5/6 * 5/6 = 125/216

Now we need to subtract this from one. The prob that at least one of the rolls will be a 3 includes every outcome, except when three consecutive non-3s are rolled. Therefore:

1- 125/216 = 91/216 = the prob that at least one 3 will be rolled

(I have illustrated the idea using three rolls) The question above involves four. I thought it was three and i don't want to go back and change it The same concept can be used though.
Hope this helps!

Simplify this further for the lack of A/C's.. = (5^4 + 4*5^3)/6^4 = (5*5^3 + 4*5^3)/6^4 = (5+4)*5^3/6^4 = 9*5^3//(3*2)^4 = 3^2*5^3/3^4*2^4 = 5^3/3^2*2^4 = 125/9*16 = 125/144 very high probability indeed

I am getting 125/216. Add p(xxxx) and p(one 3)
125/216