Three dice are rolled. What is the probability of obtaining at least one 6? The approach in the text is using the "not" rule, where we first find the probability of getting zero number of 6's. For one dice this probability is $\frac{5}{6}$ and for three die it becomes $\frac{5}{6}^3$. Hence using the not rule the probability of getting at least one 6 is $1-\frac{125}{216} = \frac{91}{216}$.
But I have a different approach which leads to a different answer - 
Let's say we have one dice then we have just 1 way to get atleast one 6. For two die, it becomes 6 ways and for three die the answer comes to be 36 ways. Hence the probability turns out to be $\frac{36}{216}$.
What is wrong with my approach?
 A: "Let's say we have one die then we have just 1 way to get at least one 6": yes, $1$ out of $6$ outcomes.
"For two dice, it becomes 6 ways". Nope, you have $1$ way to get a double 6, and $10$ ways to get only one 6 (first die: 6, second die: 1-5; first die= 1-5, second die: 6). $1+10=11$ out of $36$ outcomes.
"for three dice the answer comes to be 36 ways". Nope, you have $1$ way to get three 6s, $15$ ways to get two 6s (three times five ways: 6, 6, 1-5; 6, 1-5, 6; 1-5, 6, 6), $75$ ways to get only one 6 (three times twentyfive ways: 6, 1-5, 1-5; 1-5, 6, 1-5; 1-5, 1-5, 6). $1+15+75=91$ out of $216$ outcomes.
Hence even not using the (easier) not rule the probability of getting at least one 6 is $\frac{91}{216}$ :)
A: That approach is wrong.
Although your answer for only one die is correct you are missing  some possibilities incase of two and three dies.
Let me tell you what you are missing.

*

*In case of second die you are missing one of the two possibilities eg (1,6),(6,1).


*Actually there are 11 cases.
$$1*5 + 5*1 + 1 =11$$


*When three die appear there is chance of 6 appears only on one or two or on all dies. You need to add all the cases. If you just multply you will miss $(*,*,6)$. (* Represent any number other than 6.)
Now for three dies in your way if you agree with corrected case of two dies there will be $$6*11=66$$ ways. Which is wrong But it could be corrected by adding missing cases
Let us add missing case
$(*,*,6)$  have 5x5 = 25 cases
Now total cases will be
$$66+25=91$$
Now probability will be $$\frac{91}{216}$$
You may want to note that the case such as $(6,6,6),(6,*,6),(*,6,6)$ had included in your way of multiplying the two die case with six.
