Probability of rejection (Sampling)

Question

Consider a plant manufacturing chips of which 10% are expected to be defective. The chips are packed 30 to a box for distribution. A sample of size 10 is drawn without replacement from each box. If more than one defective is found in the sample, the box is rejected and subjected to a complete inspection. What is the probability that a box will be rejected?

I said 10 out of 30 so that's 1/3 of the chips drawn for testing. And they have 10% chance of being defective so (1/3)*.10*30. Is that right? Sorry, I don't know where to even start with this problem so I just guessed my answer. Thanks for all the help.

Answer 1

First of all $30$ chips have to be packed. We can assume that the amount of produced chips is very large. Thus the probability that a arbitrary chip, which is input into the box, is defective is always $0.1$. Now it can be calculated what the probability is that $x$ of the $10$ packed chips are defective. The random variable ($X$) is binomial distributed with the parameter $p=0.1$ and $n=30$.

Now the converse probability comes into play.The probability that more than one chip of the sample is defected is $1$ $\texttt{minus}$ the probability that no chip is defective $\texttt{minus}$ the probability that one chip is defective. Let denote the random variable for the number of defective picked chips as $Y$. $Y$ is hypergeometric distributed where $x$ is the number of defective chips in the box and $N=30, M=x$, $n=10$.

For our purpose $Y$ can be $0$ and 1 and $X$ is between $0$ and $10$ (inclusive).

The probability that more than one of the picked chips is defective is

$1-\underbrace{\sum_{x=0}^{10}\left( \frac{{x \choose 0} \cdot {30-x \choose 10}}{30 \choose 10}\cdot {30 \choose x}\cdot 0.1^x\cdot 0.9^{30-x}\right)}_{P(Y=0)} -\underbrace{\sum_{x=1}^{10}\left( \frac{{x \choose 1} \cdot {30-x \choose 9}}{30 \choose 10}\cdot {30 \choose x}\cdot 0.1^x\cdot 0.9^{30-x}\right)}_{P(Y=1)}$

Answer 2

We assume that the probability that a chip is not defective is $9/10$. We furthermore assume that the functioning of the chips is not correlated (so the random variables are independent).

The chance that a sample of 10 chips contains 10 functioning chips is, by independence, $(9/10)^{10}$. The chance that the sample contains exactly one nonfunctioning chip is equal to $10\cdot (9/10)^9 \cdot (1/10)$, where the combinatorial factor should be noted. The chance that a box is not rejected is thus $(9/10)^{10}+(9/10)^9$, the chance it is rejected is $1-(9/10)^{10}-(9/10)^{9}\approx 0.263901071$.

Random comment: this is not far from $1-\frac{2}{e}$ as $\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n=1/e$.