I am struggling with the correctness of my solution.
This is the problem:
Product produced in a factory has a probability of 0,27 to be defective. Factory control detects a defect in a probability of 0,99 (it means it won't detect a defective product in a probability of 0,01).
Question: what is the probability, that control between 5 products will find not less than 2 defective ones?
My solution:
As I understood the question, they ask the probability of at least two items will be defective.
It means:
P(at least two) = 1 - P( no defects found) - P(one defect found)
I calculated P(no defects found) that way:
f = 0.27 (probability of producing defective part)
t = 1 - f = 0.73 (probability of producing a good part)
b1 = 0.99 (probability of detecting a defected product)
b2= 1 - b1 = 0.01 (probability of detecting a defected product)
Probabilities of P(no defects found) start there:
A0 = $t^5$ (all 5 parts are not defective)
A1 = $t^4$ * k * b2 (one defective, non found)
A2 = $t^3$ * $k^2$ * $b2^2$
A3 = $t^2$ * $k^3$ * $b2^3$ (three defective, non found)
...
A5 = $k^5$ * $b2^5$
P(one defect found) starts there:
B0 (no need to duplicate)
B1 = $t^4$ * k * b2 (one defective part, only one defect found)
B2 = $t^3$ * $k^2$ * b2 * b1 (two defective parts, only one defect found)
B3 = $t^2$ * $k^3$ * $b2^2$ * b1 (three defective parts, only one defect found)
...
B5 = $k^5$ * $b2^4$ * b1 (five defective parts, only one defect found)
So, the P(zero) is equal to A0 + A1 + ... A5 = 0.2080767583...
and P(one found) is equal to B1 + B2 + ...+ B5 = 0.07619029792
And P(at least two)= 1 - P(zero) - P(one found) = 0.7157329438
I don't know, is my solution correct and answer too, maybe there is another faster way to calculate these possibilites
I found your analysis very difficult to wade through. So, my (somewhat inferior) response will simply show how I would attack the problem.
I am making some assumptions about how the problem constraints are to be interpreted. If I have misinterpreted the problem, please advise by leaving a comment, following this answer.
I am assuming that :
The probability that a specific item is defective is $(0.27).$
Assuming that a specific item is defective, the probability of the defect being detected is then $(0.99)$.
There is no possibility of a false positive. That is, if an item is not defective, then the testing will never falsely report that the item is defective.
Therefore, the probability that any item is reported as defective is $(0.27) \times (0.99).$
Testing of a specific item for a defect is independent of testing a separate item for a defect. Therefore (for example), the probability of detecting a defect on two consecutive items is $~\displaystyle \left[ ~(0.27) \times (0.99) ~\right]^2.$
For simplicity of notation, let $~p = (0.27) \times (0.99),~$ and let $~q = (1-p).$
So, $~p~$ equals the probability that a specific item will be reported as defective, and $~q~$ equals the complementary probability that the item will not be reported as defective.
Assuming that exactly $~5~$ items are tested,
for $~k \in \{0,1,2,3,4,5\},~$
let $~f(k)~$ denote the probability that exactly $k$ items are detected as defective.
Then, the desired computation is
$$1 - f(0) - f(1).$$
Then,
$$f(0) = \binom{5}{0}p^0q^{5-0} = q^5$$
and
$$f(1) = \binom{5}{1}p^1q^{5-1} = 5pq^4.$$
$\underline{\text{Final Computation}}$
$p = (0.27) \times (0.99).$
$q = 1-p.$
The probability of at least two defects being detected, out of five items tested, is
$$1 - q^5 - 5pq^4.$$