# Binomial probability

Suppose that the probability of a company supplying a defective product is $a$ and the probability that the supplied product is not defective is $b$. Before each product supplied is released for further use, it must be screened through a system which detects the potential defectiveness of the product. The probability that this process successfully detects a defective product is $x$, and the probability that it does not is $y$. What is the probability that at least one defective product will pass through the system undetected after 20 products have been screened?

i.e. $a = \text{defective product}, b = \text{non-defective product}, x = \text{screening success}, y = \text{screening fail}$

The total probability can be expressed as ${[(a + b)(x + y)]}^{20}$. However, is the probability of at least one defective product represented by $k_1 (ay) + k_2 (ay)^2 + k_{3} (ay)^3 + \ldots +k_{19}(ay)^{19} + k_{20} (ay)^{20}$ where $k_{i}~, ~i \in \mathbb{Z}$ is a coefficient of the corresponding term? I'm also interested to know if there's a simple method for computing that expression, or a more efficient method of solving this question in general.

Note: This is NOT a homework problem.

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I take it that working products are never detected as defective (so the defective-detector does not yield false positives)? –  Daan Michiels May 7 '12 at 12:53
Yes, I think that's correct. –  j_z May 7 '12 at 12:55

Call $A_n$ the event that the $n$th product is defective and undetected, then $\mathrm P(A_n)=ay$. The event $B_k$ that at least one defective product will pass through the system undetected after $k$ products have been screened is the union of $k$ events $A_n$ hence the complement of $B_k$ is the intersection of their complements. Although this necessary hypothesis is not mentioned, one probably assumes that the events $(A_n)_n$ are independent, then the complementary events are i.i.d. with probabilities $1-ay$. Thus, $\mathrm P(B_k)=1-(1-ay)^k$.
Do you confirm that $a+b=x+y=1$? –  Did May 7 '12 at 13:05
If the $n$th product is defective and undetected, then doesn't $P(A_n)=ay$? –  j_z May 7 '12 at 18:13