Factories A and B produce computers. Factory A produces 3 times as many computers as factory B. The probability that an item produced by factory A is defective is 0.03 and the probability that an item produced by factory B is defective is 0.045. A computer is selected at random and it is found to be defective. What is the probability it came from factory A?

P(A∣D) = 0.667

Step-by-step explanation:

We are given;

P(A) = 3P(B)

P(D|A) = 0.03

P(D|B) = 0.045

Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.

Using Bayes' Rule and Law of Total Probability, we will get;

P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]

Plugging in the relevant values, we have;

P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]

P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]

P(B) will cancel out to give;

P(A∣D) = 0.09/0.135

P(A∣D) = 0.667

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.