Question

A quality control inspector at the store keeps looking at randomly selected cartons of eggs until he finds one with at least 2 broken eggs. Find the probability that this happens in one of the first three cartons he inspects.

Answer 1

The question is a probability problem where the student is looking for the likelihood of finding a carton with at least 2 broken eggs in the first three inspections, which can be solved using the geometric or negative binomial distributions, but specific probability values are missing.

Step-by-step explanation:

The question the student asked relates to a scenario where a quality control inspector selects cartons of eggs looking for one with at least 2 broken eggs within the first three cartons inspected. This is a problem that can be solved using probability concepts, specifically the geometric or negative binomial distributions, depending on whether the probability of finding such a carton in any given inspection is known or not. Unfortunately, the exact probability of a carton having at least 2 broken eggs is not provided, which prevents a complete answer from being given. However, if such a probability (say p) were known, then the probability of finding such a carton on the first, second, or third inspection would be calculated as p(1-p)^0 + p(1-p)^1 + p(1-p)^2.

Answer 2

To find the probability of finding a carton with at least 2 broken eggs in the first three cartons inspected, we can use the concept of complementary probability. The probability is approximately 0.0408.

Step-by-step explanation:

To find the probability that the quality control inspector finds a carton with at least 2 broken eggs in the first three cartons inspected, we can use the concept of complementary probability. First, let's find the probability that none of the three cartons have at least 2 broken eggs. To do this, we need to find the probability that each carton has 0 or 1 broken egg:

1. In the first carton, the probability of 0 or 1 broken egg is (10/144) + (12/144) = 22/144 = 11/72.
2. In the second carton, the probability of 0 or 1 broken egg given that the first carton had 0 or 1 broken egg is (10/143) + (12/143) = 22/143.
3. In the third carton, the probability of 0 or 1 broken egg given that the first two cartons had 0 or 1 broken eggs is (10/142) + (12/142) = 22/142 = 11/71.

Now, we can find the probability that at least 2 broken eggs are found in the first three cartons by subtracting the probability of none of the cartons having 2 broken eggs from 1:

Probability at least 2 broken eggs in first three cartons = 1 - (11/72) * (22/143) * (11/71) = 101/2478 ≈ 0.0408 (rounded to 4 decimal places).