Question

If probability that type 1 is defect is P(A)=0.02, and the probability that type 2 is defect is P(B)=0.06 and P(A) and P(B) are independent.

P(C) is the event that it is defective. What is P(C | A)?

I know P(C) is 0.0788, because P(C) = P(A) + P(B) - P(A)P(B)

But how do I find P(C | A)? I don't get it.

UPDATE: Nvm, it's just 1. I'm stupid. I'll give a thumbs up to the first one who gives a random answer

Answer #1

We are given P(A) = 0.02, P(B) = 0.06

Since A and B are independent:

P(A and B) = P(A) * P(B) = 0.02 * 0.06 = 0.0012

C is the event that it is defective. Therefore:

P(C) = P(A or B) = P(A) + P(B) - P(A and B)

= 0.02 + 0.06 - 0.0012

= 0.0788

P(C|A) = P(A and C)/P(A)

= P(A)/P(A) ...because P(A and C) = P(A and (A or B)) = P(A)

=
**1**

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