In a
manufacturing process, each item leaving an assembly line will be examined by
two inspectors: A and B individually. Based on past experiences, Inspector A detects 60% of the defective items. If the defective items are not detected by Inspector A, 90% of them will be detected by Inspector B. However, 60% of the defective items are not detected by Inspector B given that they are detected by Inspector A.
What is the probability that a defective item is detected by Inspector A, given
that it is detected by Inspector B?
To find the probability that a defective item is detected by Inspector A given that it is detected by Inspector B, we can use Bayes' theorem.
Let's assume:
A = Event that a defective item is detected by Inspector A
B = Event that a defective item is detected by Inspector B
We want to find P(A | B), i.e., the probability that A occurs given that B occurs.
From the given information:
P(A) = 0.60 (probability that Inspector A detects a defective item)
P(B | A) = 0.40 (probability that Inspector B detects a defective item given that it is detected by Inspector A)
By Bayes' theorem:
P(A | B) = (P(B | A) * P(A)) / P(B)
To find P(B), we need to consider two cases:
1. Defective items detected by both Inspector A and Inspector B.
2. Defective items detected only by Inspector B.
Let's calculate P(B) step-by-step:
1. Defective items detected by both A and B:
Here, Inspector A detects 60% of defective items and Inspector B detects 90% of non-detected items by Inspector A.
So, the probability of defective items detected by both A and B is:
P(Defective items detected by both A and B) = (0.60 * 0.10) = 0.06
2. Defective items detected only by B:
Here, Inspector A detects 60% of defective items and Inspector B detects 60% of non-detected items by Inspector A.
So, the probability of defective items detected only by B is:
P(Defective items detected only by B) = (0.40 * 0.60) = 0.24
Therefore, P(B) = P(Defective items detected by both A and B) + P(Defective items detected only by B)
= 0.06 + 0.24
= 0.30
Now, substituting the values into Bayes' theorem:
P(A | B) = (P(B | A) * P(A)) / P(B)
= (0.40 * 0.60) / 0.30
= 0.24 / 0.30
= 0.80
Therefore, the probability that a defective item is detected by Inspector A, given that it is detected by Inspector B, is 0.80 or 80%.
To find the probability that a defective item is detected by Inspector A, given that it is detected by Inspector B, we can use Bayes' Theorem.
Let's break down the information provided:
- The probability of Inspector A detecting a defective item is 60% (0.6).
- The probability of an item not being detected by Inspector A and then being detected by Inspector B is 90% (0.9).
- The probability of an item being detected by Inspector A and not by Inspector B is 60% (0.6).
Now we can apply Bayes' Theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
Where:
P(A|B) is the probability of detecting by Inspector A given that it is detected by Inspector B.
P(B|A) is the probability of being detected by Inspector B given that it is detected by Inspector A.
P(A) is the probability of detecting by Inspector A without considering Inspector B.
P(B) is the probability of detecting by Inspector B without considering Inspector A.
Let's substitute the values:
P(A|B) = (0.9 * 0.6) / (0.9 * 0.6 + 0.1 * 0.4)
Simplifying further:
P(A|B) = 0.54 / (0.54 + 0.04)
P(A|B) = 0.54 / 0.58
The probability that a defective item is detected by Inspector A, given it is detected by Inspector B, is approximately 0.931 or 93.1%.
Therefore, there is a 93.1% chance that an item detected by Inspector B was also detected by Inspector A.