Find the probability of success in the first least 20 consecutive attempts The battery of a particular car brand starts with probability of $0.95$. Find the probability that the battery starts on the first at least $20$ consecutive attempts.
"success": the battery starts
"failure": the battery doesn't start
I have some issues with interpreting the at least part. If the question was given without it, then the answer would be the probability of success in the first attempt and success in the second attempt and ... and success in the 20th attempt. However, I managed to get to an answer, but I'm not completely sure about its validity: 1 - probability of failure in the first attempt and failure in the second attempt and ... and failure in the 19th attempt.
Thank you in advance!
 A: Let us define success as start and failure as non-start. The probability that you have n consecutive successes before a failure follows geometric distribution as below:
$P(X=k) = p^kq$, $k = 0,1,2,..$where $p$ is the probability of start and $ q$ is the probability of non start and k is the number of starts.  What you have been asked is to find the probability of atleast $20$ consecutive starts before a non-start.  In other words, 
$P(X\ge 20) $
$$=1-P(X\le 19) = 1- q-pq-p^2q-p^3q-\cdots - p^{19}q$$
$$ 1- q(1+p+p^2+..p^{19})$$ $$= 1-q\frac{(1-p^{20})}{1-p} = 1-1+p^{20} = p^{20}=0.358486$$ where $p = 0.95$
A: Let $i$ be an integer between 1 and 20.
Probability that the car fails for the first time at the first try : $0,05.$
Probability that the car fails for the first time at the second try : $0,95\times 0,05$
...
Probability that the car fails for the first time at the $i$-th try : $0,95^{i-1}\times 0,05$ ($i-1$ consecutive successes followed by a failure).
This gives a partition on the opposite event.
Consequently, the probability you're looking for is :
$1-0,05\times(1+0,95+\ldots+0,95^{19})=1-0,05\frac{1-0,95^{20}}{1-0,95}=1-(1-0,95^{20})=0,95^{20}$
A: If you are looking for the probability that at least one of the first 20's attempts will make the battery start, you are looking for 1 - "all the 20's will fail". is it making it any easier for you? Am I getting your question right?
Edit:
you can take the probability for all first 20 successes (0.95^20) and multiply it in all your options, which is the geometric series $\sum_{i=0}^{\infty}(0.95)^i$
You can calculate where this series goes to. 
