A certain system can experience three different types of defects. Let A_i (i = 1

Question

A certain system can experience three different types of defects. Let A_i (i = 1, 2, 3) denote the event that the system has a defect of type i. Suppose that the following probabilities are true. P(A_1) = 0.10 P(A_2) = 0.07 P(A_3) = 0.06 P(A_1 union A_2) = 0.11 P(A_1 union A_3) = 0.13 P(A_2 union A_3) = 0.11 P(A_1 intersection A_2 intersection A_3) = 0.01 Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? (Round your answer to four decimal places.) 83 Given that the system has a type 1 defect, what is the probability that it has all three types of defects? (Round your answer to four decimal places.) _________ Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? (Round your answer to four decimal places.) __________ Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect? (Round your answer to four decimal places.) __________

Explanation / Answer

here P(A1nA2)=P(A1)+P(A2)-P(A1UA2|)=0.1+0.07-0.11=0.06

P(A1nA3)=0.1+0.06-0.13=0.03

P(A2nA3)=0.07+0.06+-0.11=0.02

a) hence P(A2|A1)=P(A1nA2)/P(A1)=0.06/0.1=0.6

b)P(A1nA2nA3|A1) =P(A1nA2nA3)/P(A1)=0.01/0.1=0.1

c)probabilty of one type of defect =P(A1UA2UA3)

=P(A1)+P(A2)+P(A3)-P(A1nA2)-P(A2nA3)-P(A1nA3)+P(A1nA2nA3)

=0.1+0.07+0.06-0.06-0.03-0.02+0.01=0.13

and probability of exaclty one type of defect=

P(A1)+P(A2)+P(A3)-(2*P(A1nA2)+2*P(A2nA3)+2*P(A1nA3)-3P(A1nA2nA3))

=0.1+0.07+0.06-(2*0.06+2*0.03+2*0.02-3*0.01)=0.04

hence required probability =0.04/0.13=0.3077

d)asprobability of both type of defects = P(A1nA2)=0.06

and probability of not having 3 rd defect while having first two =P(A1nA2)-P(A1nA2nA3)=0.06-0.01=0.05

hence probability=0.05/0.06=0.8333

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