Binomial Probability Word Problem. Hirsbrunner produces tubas and ships them in lots of twenty. Suppose that 60% of all such lots contain no defective tubas, 30% contain one defective, and 10% contain two defectives. Now suppose that a lot is inspected, with two tubas being selected from it at random, and neither is found to be defective.

engausidarb

engausidarb

Answered question

2022-09-11

Binomial Probability Word Problem
This problem appears very simple, but I am almost positive that it should not be so simple. For the question below I reason that the probability for part a is 0.6 and part b is 0.6 as well. What am I missing?
Hirsbrunner produces tubas and ships them in lots of twenty. Suppose that 60% of all such lots contain no defective tubas, 30% contain one defective, and 10% contain two defectives. Now suppose that a lot is inspected, with two tubas being selected from it at random, and neither is found to be defective.
a) What is the probability that there are no defectives in that lot?
b) Suppose that the inspected lot is from a shipping container that contains 10 lots, and the other 9 lots were not inspected. What is the probability that there are no defectives in that container?

Answer & Explanation

Adolfo Lee

Adolfo Lee

Beginner2022-09-12Added 17 answers

Step 1
For a, you are missing that not finding a defective improves the odds that there is none. The chance of finding no defectives is 60 % ( no defectives ) + 30 % ( 19 2 ) ( 20 2 ) ( one defective and missed it ) + 10 % ( 18 2 ) ( 20 2 ) ( two defective and missed them ) 60 % + 27 % + 8.05 % = 95.05 %.
Step 2
The chance that there are no defectives given the inspection is then 60 95.05 63.12 %
For b, if you don't inspect any others, the chance that the other nine lots have no defectives is 0.6 9 and you need to multiply the answer from a by this.
potrefilizx

potrefilizx

Beginner2022-09-13Added 2 answers

Step 1
(a) Let G be the event that the whole lot is good (zero defectives), and let P be the event there were no defectives in the sample of 2. We are asked for the conditional probability Pr(G|P), the probability that the whole lot is good given the information that the sample of 2 had no defectives. By a formula which is likely familiar to you, essentially the definition of conditional probability, we have
Pr ( G | P ) = Pr ( G P Pr ( P ) .
It remains to find the probabilities on the right.
We go first for the harder one, Pr(P). The event P can happen in three ways: (i) the chosen lot has no defectives, and (of course) no defectives are found; (ii) the chosen lot has one defective, and no defectives are found; or (iii) the chosen lot has two defectives, and no defectives are found.
The probability of (i) is clearly 0.6. Note that this is also Pr ( P G ).
The probability that the chosen lot has 1 defective is 0.3. If we sample from this lot, then with probability 19 20 the first tested item is OK, and given that happened, the probability the second item tested is OK is 18 19 , for a product of 18 19 . So the probability of (ii) is (0.3)(18/20), which is exactly 0.27.
Step 2
In the same way, we find that the probability of (iii) is (0.1)(18/20)(17/19), which is approximately 0.0805.
So Pr ( P ) 0.6 + 0.27 + 0.0805, which is about 0.9505.
Finally, for our conditional probability, divide Pr ( G P ) by Pr(P). We get about 0.63123.
(b) We are expected to assume independence. For the inspected lot, the probability it has no defectives is approximately 0.63213. The 9 uninspected lots each have probability 0.6 of having no defectives. Multiply.

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