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Q. The probability that a product is defective is 0.01. The products are packed in boxes; each box contains 10 products. A company orders a consignment of 12 boxes of products. A purchaser randomly opens three boxes, and accepts the consignment if the 3 boxes altogether contain at most one defective product. Calculate the probability that the purchaser rejects the consignment. Round your answer to 3 decimal places.

My Sol:

I understand the distribution is binomial but I am not sure of the full solution for this. I am thinking that n=10 (each box contains 10 products) and p=0.99 (non-defective). But I am not sure how to bring in the 3 out of the 12 boxes to find the probability...any help is appreciated. Thank you.

eun ji
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    "I am not sure how to bring in the 3 out of the 12 boxes" : I think that Jellyfish's answer is dead on accurate. That is, regardless of which 3 boxes are selected, there will be $(3 \times 10)$ products to separately examine. Let $P_0$ denote the probability that exactly $(0)$ of the 30 products are defective, and let $P_1$ denote the probability that exactly $(1)$ of the 30 products are defective. Then the desired computation is $1 - (P_0 + P_1)$. This is exactly what Jellyfish's answer computes. – user2661923 Jul 19 '21 at 03:00

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Each product in each box has a probability of being defective of $.01$, and there are three boxes of $10$ products examined. Since the boxes are all the same it doesn't matter if there are 12 boxes total, 100 boxes, or 3 boxes in their inventory. Thus the number of defective products in all three boxes is $X\sim\text{Binomial}(30, .01)$. The question asks for

$$\begin{split}1-[P(X=0)+P(X=1)]&=1-[.99^{30}+30(.01)(.99)^{29}]\\ &=1-.9639\\ &=.036\end{split}$$

Vons
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    +1 : see also my comment following the question. My one criticism is that in the second line, you might use \approx, unless the answer actually is exactly $(1 - 0.9639)$. Ditto the 3rd line. – user2661923 Jul 19 '21 at 03:03