Probability that on three rolls of dice, there will be at least one 6 showing up? 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 ?
 A: Throw the dice one at a time, so that you have a "first", "second" and "third" die.
Your first part $\frac{1}{6}\frac{5}{6}\frac{5}{6}$ is not the chance of getting one six.  It is the chance of the first die being six while the other two are not.  You also need to add the chances that the second or third die is six and the other two not.  Since these are all the same, you just need to multiply by three.
Your second part $\frac{1}{6}\frac{1}{6}\frac{5}{6}$ also needs to be tripled for the same reason. In this case there are three dice that may be not-six.
Your last part is fine since there is only one way all three dice can be six.
Adding it all together you get $3\frac{1}{6}\frac{5}{6}\frac{5}{6}+3\frac{1}{6}\frac{1}{6}\frac{5}{6}+\frac{1}{6}\frac{1}{6}\frac{1}{6} = \frac{91}{216}$
This calculation is relatively painless with three dice but quickly gets worse as you add dice.
However, there is an easier way.  Turn the question around and ask what is the probability that no die shows six.  This is clearly $(\frac{5}{6})^3=\frac{125}{216}$
So, the probability that some die shows six is $1-\frac{125}{216}=\frac{91}{216}$
A: You have forgotten to include th fact that any six can occur in any of the rolls, i.e. you have multiply the probabilities for exactly one and two sixs by three, giving
$$P=3\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}+3\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}=\frac{91}{216}.$$
It may be easier to compute the complementary probability of no six showing up: $$P^c=\left(\frac{5}{6}\right)^3=\frac{125}{216}$$
and using $$P=1-P^c=\frac{216-125}{216}=\frac{91}{216}$$
A: Case I : one six .. the six can show up at first, second, or third. And
similiarly for other cases,
So your answer should be
$$\binom{3}{1} (1/6) (5/6).(5/6)+\binom{3}{2} (1/6).(1/6).(5/6)+\binom{3}{3}(1/6).(1/6).(1/6)$$
or you could do $$1-(5/6)^3$$
A: 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 $1 - (5/6)^3 = 91/216$
