Statistics and Probabilities- Distributions A quality control engineer tests the quality of produced computers. Suppose that 5% of computers have defects and defects occur independently of each other. I need to find the probability that the engineer has to test at least 5 computers in order to find 2 defective ones.
I thought of it and came up with a solution though I'm not sure whether it's correct or not. 

The probability of finding $2$ defective is $p= 0.05*0.05=0.025$
  $P(T\geq 5)= 1- [ P(T=4)+P(T=3)+(T=2)]= \\ 1-[0.025*(0.975)^3+0.025*(0.975)^2+0.025*(0.975)^1]=\\
1-[0.025*0.975(1+0.975+(0.975)^2)= 0.9300442$

 A: Let $X$ be the number of defectives in the first $4$ computers tested. We want the probability that $X=0$ or $X=1$. Note that $X$ has binomial distribution. We have $\Pr(X=0)=0.95^4$ and $\Pr(X=1)=\binom{4}{1}(0.05)(0.95)^3$. Add.
Remark: The approach taken in the OP is correct. albeit a little longer. We do the details for comparison. Let $Y$ be the number of trials (computers) until the second bad. We want $\Pr(Y\ge 5)$. 
We go after the probability of the complementary event. So we compute $\Pr(Y=2)+\Pr(Y=3)+\Pr(Y=4)$.
Clearly $\Pr(Y=2)=(0.05)^2$. 
For $Y=3$ we must have one bad in the first two trials, then a bad. The probability is $\binom{2}{1}(0.05)(0.95)(0.05)=\binom{2}{1}(0.95)(0.05)^2$.
Similarly, to have $Y=4$ we need exactly one bad in the first three trials, then a bad. The probability is $\binom{3}{1}(0.95)^2(0.05)^2$. 
Add up, subtract from $1$. We get about $0.985983$.  
A: I saw the question and answer from one of the website about the b) from https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.408149.html
Note: The probability of x successes in n trials is: 
In this case p = .05 & q = .95 
a) P(exactly 3 of 20 being defective computers) = 1140(.05)^3(.95)17 = .0596
(b) P( test at least 5 computers to find 2 defective ones)
P = 1 - (.05)^0*(.95)^5 + 5*(.05)^1*(.95)^4 = .0226
