Calculating probabilities of binomial distributions for two methods

Question

I'm struggling with this question on binomial distributions:

A factory is considering two methods, I and II, of checking the quality of production of the batches of items it produces.

Method I consists of taking a random sample of 10 items from a large batch and accepting the batch if there are no defectives and rejecting the batch if there are two or more defectives. If there is one defective in the sample, another random sample of 10 items is taken from the batch. The batch is accepted if there are no defectives in this second sample and rejected otherwise. Method II consists of taking a random sample of 20 items from a batch and accepting the batch if there is at most one defective in the sample. Otherwise the batch is rejected.

The factory knows that 1% of the items it produces are defective and wishes to use that method of checking the quality of production for which the probability of accepting a batch is larger. Decide whether the factory should use Method I or Method II.

The first thing I tried to do is categorise by the number of samples needed, along with their probabilities. For Method 1, there are either 1 sample (initially accepted or rejected) or 2 samples (if it needs to be re-tested). Since there are the chance for 0-10 defections, I assumed the denominator is 11:

$$ \begin{align} P(S=1) & =\binom{10}{1} \left(\frac{1}{11} + \frac{9}{11}\right) = \frac{100}{11} \\ P(S=2) & =\binom{10}{2} \left(\frac{1}{11}\right) = \frac{45}{11} \\ \end{align} $$

But the probability for Method I is $0.987$ and Method II is $0.983$. Can someone explain how these numbers are calculated?

Answer 1

Remember that $\binom{n}k$ essentially gives us the number of ways to take k items at a time from a set of n. Or likewise, if we have a set of n items where k are of one type and n-k are of another type, how many unique ways we can arrange them in a line. This is helpful for probability problems like this, because otherwise, trying to figure out "What's the probability that 2 out of these 20 items are defective" might have you thinking, "Well I need the probability of only items #1 and #2 being defective, plus the probability of only items #1 and #3 being defective, plus the probability of...(all the other possible combinations, which in this case is a total of 190)." Instead, you can just calculate the probabiltiy for one example, and multiply it by the number of ways it can happen. So I don't really understand your rationale for the "11" denominator. None of the probabilities in the problem are something out of 11.

Since "The factory knows that 1% of the items it produces are defective", then there's a probability of 0.99 of any single sample being non-defective. With Method I, one of two things happens for acceptance: A) the first sample of 10 items has no defects, or B) the first sample has exactly one defect, and the next sample of 10 has no defect. The first probability is just 0.99¹⁰. The probability of the first same having exactly one defect is $\binom{10}1$(0.99⁹)(0.01¹). That is, the number of ways one defect can show up in a sample of 10, times the probability of 9 items being not defective, and one being defective. But we said "and the next sample of 10 has no defect", so we have to multiply this by another 0.99¹⁰.

Adding up scenarios A and B, we get:

0.99¹⁰ + [ $\binom{10}1$(0.99⁹)(0.01¹) x 0.99¹⁰ ] = 0.98699...

For Method II, one of two things happens for acceptance: 1) No defects in the sample of 20, or 2) exactly one defect in the sample of 20. If you use the same sort of strategy as above, you should get a result that's approximately 0.9831.

Answer 2

$P(S=1) = 0.99^{10} + \binom{10}{1}0.99^9\times0.01\times 0.99^{10}=0.9870$

$P(S=2) = 0.99^{20} + \binom{20}{1}0.99^{19}\times 0.01=0.9831$