Acomputer manufacturer uses chips from three sources. chips from sources a, b, and c are defective with probabilities 0.005, 0.001, and 0.010, respectively. you can assume that the proportions of chips from a, b and c are 0.5, 0.1, and 0.4 respectively. if a randomly selected chip is found to be defective, find the probability that the manufacturer was a and the probability that the manufacturer was c.

Step-by-step explanation:

A="The chip is from the manufacturer A"

B="The chip is from the manufacturer B"

C="The chip is from the manufacturer C"

D="The chip is defective"

P(A)=0.5

P(B)=0.1

P(C)=0.4

P(D|A)=0.005

P(D|B)=0.001

P(D|C)=0.01

P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)=0.005*0.5+0.001*0.1+0.01*0.4

P(D)=6.6x10-3

Based on Bayes rule:

P(A|D)=P(D|A)P(A)/P(D)=0.005*0.5/6.6x10-3=0.38

P(C|D)=P(D|C)P(C)/P(D)=0.01*0.4/6.6x10-3=0.60

i believe the answer to be 48

step-by-step explanation:

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.