Let's say I want to know what's the probability of rolling two 6-sided dice and getting a 6 in at least one of them.
The probability of not getting a 6 in one die is 3/6. I know I can use the following formula for that:
Probability of not getting a 6: (5/6)^2 -> 25/36
- Formula: (probability of not getting a 6)^(number of tries)
Probability of getting a 6: 1-(5/6)^2 -> 11/36
- Formula: 1-(probability of not getting a 6)^(number of tries) All good until this point.
This made me think: To calculate the probability of "getting a 6 in at least one die", I need to use the probability of "not getting a 6 in at least one die" and subtract that from the "total probability".
Which leads to my question: Is there a general formula that lets me calculate the probability of "getting a 6 in at least one die" that doesn't rely on the subtracting the opposite probability from one?
P.S.: I'm not asking whether there's a "better" formula, just wondering if there's a different general way of calculating it (even if it's more complicated or longer).
Let $p_0$ be the probability of getting a 6 in each individual throw, let $n$ be the number of dice throws you carry out.
To calculate the probability of getting atleast one 6, you would need to sum the probabilities of getting exactly $i$ 6's, where $i$ ranges from 1 to $n$.
$$P\text{(atleast one six)=}\sum_{i=1}^n P(\text{exactly }i \text{ sixes)}$$
To find $P(\text{exactly }i \text{ sixes)}$, you would need to get a six in any $i$ trials out of the $n$ trials, and one of the other five numbers in the rest of the ($n-i$) trials. Keep in mind that the sixes can be in any order, so there are a total of $n\choose i$ possible combinations. $$\implies P(\text{exactly }i \text{ sixes)}= (p_0)^i \cdot(1-p_0)^{n-i}\cdot {n\choose i} $$
$$\implies P\text{(atleast one six)=}\sum_{i=1}^n \left[(p_0)^i \cdot(1-p_0)^{n-i}\cdot {n\choose i} \right] $$
At this point, you could get your answer by calculating this sum manually, but a little manipulation can make this expression much simpler. Notice that $$\sum_{i=0}^n \left[(p_0)^i \cdot(1-p_0)^{n-i}\cdot {n\choose i} \right]=(1-p_0)^n+\sum_{i=1}^n \left[(p_0)^i \cdot(1-p_0)^{n-i}\cdot {n\choose i} \right]$$
By the binomial theorem, the left hand side is simply equal to $(p_o+1-p_0)^n=1 $ $$\implies \sum_{i=1}^n \left[(p_0)^i \cdot(1-p_0)^{n-i}\cdot {n\choose i} \right] =1-(1-p_0)^n$$
Recall that $(1-p_0)$ is simply the probability of not getting a six in a single trial, and thus $(1-p_0)^n$ is the probability of not getting a six in $n$ trials. So you end up with the same simplified expression.