Let's say we're talking about rolling a six. Here's what I know...
For one roll ...
(probability of rolling a six) = 1/6
(probability of NOT rolling a six) = 5/6
(probability of rolling a six) + (probability of NOT rolling a six) = 1/6 + 5/6 = 1
For 'x' number of rolls ...
(probability of rolling a six 'x' times) = (1/6)^x
(probability of NOT rolling a six 'x' times) = (5/6)^x
(probability of rolling at least one six in 'x' rolls) = 1 − (probability of NOT rolling a six 'x' times) = 1 − (5/6)^x
My question:
When trying to calculate the probability of rolling at least one six in 'x' rolls (the last line above), is there an alternative to this 1 − p(NOT six) method? Isn't the probability of rolling at least one six in 'x' rolls somehow a combination of the probabilities of rolling one six, two sixes, three sixes, four sixes etc.?
Probablity to get "at least" on six in $n$ rolls is obviously
$$P(X=1)+P(X=2)+\dots P(X=n)=1-P(X=0)$$
to calculate this probability you can use the binomial distribution that is
$$P(X=k)=\binom{n}{k}p^k\cdot(1-p)^{n-k}$$
Let's have a simple example:
Let's roll 3 times a die (or one time 3 dice) and calcualate the probability to have "at least one six"
First approach (direct approach)
$$P(X\geq 1)=\binom{3}{1}\cdot\left(\frac{1}{6}\right)^1\cdot\left(\frac{5}{6}\right)^2+\binom{3}{2}\cdot\left(\frac{1}{6}\right)^2\cdot\left(\frac{5}{6}\right)^1+\binom{3}{3}\cdot\left(\frac{1}{6}\right)^3\cdot\left(\frac{5}{6}\right)^0=\frac{91}{216}$$
Second (indirect) approach
$$P(X\geq 1)=1-P(X=0)=1-\binom{3}{0}\cdot\left(\frac{1}{6}\right)^0\cdot\left(\frac{5}{6}\right)^3=\frac{91}{216}$$