A certain random attempt is carried out $n$ times ($n \in \mathbb{N}, \ n \geq 2$). With every implementation occurs "Success" (regardless of the other implementations) with a probability of $0.2$.
The following applies: $n = 10$.
On the tenth execution (and only this one) increases due to changed conditions the probability of "success" occurring is $0.3$.
- Calculate the probability that “success” occurs exactly once.
Do we use the formula $$P(X=k)=\binom{n}{k}\cdot p^k\cdot (1-p)^{n-k}$$ for the first $9$ attemps and with $p=0.2$ plus the term for the last attempt with $p=0.3$ ?
So do we have then $$P(X=1)=\binom{9}{1}\cdot 0.2^1\cdot (1-0.2)^{9-1}+\binom{1}{1}\cdot 0.3^1\cdot (1-0.3)^{1-1}=9\cdot 0.2\cdot 0.8^{8}+ 0.3$$ ? Or how do we calculate that probability?
It should be,
$ \small \binom{9}{1}\times 0.2 \times (1-0.2)^8 \times \color {blue } {(1 - 0.3)} + \color {blue } {(1-0.2)^9} \times 0.3$
See the additions marked in blue.
When one of the first nine attempts is a success then the last attempt has to be a failure. That is why we multiply the first term also by $(1 - 0.3)$.
Similarly if the last attempt is a success, we must have all previous nine attempts as failure. So we multiply the second term by $(1 - 0.2)^9$.