Products produced in a certain factory are inspected against three standards A, B, and C to determine pass or fail.  From the long experience of this plant, we know that 1% of A is defective, 3% of B, and 2% of C.  Calculate the probability that one product will pass if it must pass standard A and if it passes either standard B or C, then this product is recognized as a passing product.  However, those that pass standards A, B, and C are said to be independent of each other.

이거 질문넣었음 수학의정석 확통 연습문제 7-23번

한국어 : 어느 공장에서 생산되는 제품은 3개의 규격 A, B, C에 관하여 검사해서 합격, 불합격을 결정한다. 이 공장의 오랜 경험으로부터 A에 관하여는 1%, B에 관하여는 3%, C에 관하여는 2%의 불량품이 나온다는 것을 알고 있다. 규격 A에는 반드시 합격해야 하고, 규격 B, C에는 어느 쪽인가에 합격하면 이 제품은 합격품으로 인정한다고 할 때, 한 개의 제품이 합격할 확률을 구하여라. 단, 규격 A, B, C에 합격하는 것은 서로 독립이라고 한다



답변 : Let's define the following events:

A: the product passes standard A
B: the product passes standard B
C: the product passes standard C
P: the product passes overall (according to the given criteria)
We want to calculate P(P|A ∩ (B ∪ C)), which is the probability that the product passes overall given that it passes standard A and passes either standard B or C.

We can use the law of total probability to express P(P|A ∩ (B ∪ C)) in terms of conditional probabilities:

P(P|A ∩ (B ∪ C)) = P(P ∩ A ∩ (B ∪ C)) / P(A ∩ (B ∪ C))

To calculate the numerator, we can use the product rule of probability:

P(P ∩ A ∩ (B ∪ C)) = P(P | A ∩ (B ∪ C)) * P(A) * P(B ∪ C | A)

We are given that those products that pass either standard B or C are recognized as passing products, so we can write:

P(B ∪ C | A) = P(B ∪ C | A ∩ P)

Since A, B, and C are said to be independent of each other, we can use the product rule to express this conditional probability:

P(B ∪ C | A ∩ P) = P(B | A ∩ P) + P(C | A ∩ P) - P(B ∩ C | A ∩ P)

We are not given the probabilities of B and C given that the product passes overall, but we are given the probabilities of defective products for each standard. Since a product that is not defective passes, we can use the complements of these probabilities to represent the probabilities of passing:

P(passing | A) = 1 - 0.01 = 0.99 (1% of A is defective)
P(passing | B) = 1 - 0.03 = 0.97 (3% of B is defective)
P(passing | C) = 1 - 0.02 = 0.98 (2% of C is defective)
To calculate P(P | A), we can use the law of total probability again:

P(P | A) = P(P ∩ A) / P(A)
= P(P | A) * P(A | P) / P(A)

Since a product that passes standard A and either standard B or C is recognized as passing, we have:

P(P | A) = P(B ∪ C | A)
= P(B | A) + P(C | A) - P(B ∩ C | A)

Using the probabilities of passing and the fact that A, B, and C are independent, we can calculate:

P(B | A) = P(passing | B) * P(B) = 0.97 * 0.99 = 0.9603
P(C | A) = P(passing | C) * P(C) = 0.98 * 0.99 = 0.9702
P(B ∩ C | A) = P(B | A) * P(C | A) = 0.9603 * 0.9702 = 0.9324
Now we can substitute these values into the expressions we derived above:

P(B ∪ C | A ∩ P) = P(B | A ∩ P) + P(C | A ∩ P) - P(B ∩ C | A ∩ P)
= 0.9603 + 0.9702 - 0.9324


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