What is the probability that on three rolls of dice, there will be at least one 6 showing up?
Attempt:
Since there can be one six or two sixes or three sixes on three rolls, I considered separate cases and added them up.
So $(1/6)(5/6)(5/6) + (1/6)(1/6)(5/6) + (1/6)(1/6)(1/6) = 31/216$, but answer is incorrect as per book. Can anyone suggest where I am wrong ?
Best Answer
Your answer adds three cases - 6xx, 66x and 666 (where x is anything non-six). You've omitted x6x, xx6, x66, 6x6.
The easiest way to do it is to see that you fail only when you roll 3 non-sixes, which happens with probability $(5/6)^3$ - so you succeed with probability $1 - (5/6)^3 = 91/216$