# After $10$ inspections with no defect, whats the prob of that number of inspection is no more than $20$?

I have a problem which I'm not sure how to solve. It goes as follows:

A production line has a $5$% defective rate, and its products are inspected one by one until the first defect is found. Given that the first $10$ inspections do not find any defect, what is the probability that the number of inspections is no more than $20$?

I tried the following approach:

Given that this is a Bernoulli process, I thought that if there were no defective parts encountered in the first 10 inspected products, the next 10 products (till $X_{20}$) will be independent from the past, which means a new Bernoulli process of 10 trials. So, if $S$ is the sum of each $X_i$, then I could calculate its PMF with parameters $p=.05$ and $t=10$, where $t$ is the number of trials:

$p_s(1) = \binom{10}{1}(0.05)(0.95)^9$

$p_s(1) = .3151$

Yet this is wrong. I'm not sure if I had to use a conditional approach (because the second round of inspection occurs in the conditional universe where there were no defects in the first $10$) so as to say that:

Event A : one is defective in $11$ through $20$

Event B : none is defective in the first $10$

$P(A \cap B) = P(A) P(B)$

In which case:

$P(A) = p_s(1) = \binom{10}{1}(0.05)(0.95)^9$

$P(B) = p_s(0) = \binom{10}{0}(0.95)^{10}$

Suppose defects are iid. There are more than $20$ inspections if the next $10$ parts to be inspected are not defective. This occurs with probability $0.95^{10}$.
Hence, the probability that there are no more than $20$ inspections is $1 - 0.95^{10}$.