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

Question

A certain system can experience three different types of defects. Let A_ (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 (a) Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? 0.6 (b) Given that the system has a type 1 defect, what is the probability that it has all three types of defects? 0.1 (c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? .09091 (d) 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? 0.8333

Explanation / Answer

a. P(A1 and A2)/P(A1) = (P(A1)-P(A2)-P(A1 union A2))/P(A1) = (.10+.07-.11)/.10 = .6

b. P(A1 and A2 and A3) /P(A1) = .01/.1 = .1

c. P( one defect | atleast 1 defect) = ?

P( A1 union A2) = P(A1)+P(A2)-P(A1 int A2)
.11 = .10+.07- P(A1 int A2)
P(A1 int A2) = .06

Similarly, P(A2 int A3) = .07+.06-.11 = .02
Similarly, P(A3 int A1) = .10+.06-.13 = .03

So, P( one defect | atleast 1 defect) =(.1-.06-.03+.01+.07-.02-.06+.01+.06-.03-.02+.01) /(1-.10-.07-.06+.06+.02+.03-.01) = .04598

d. P( given (A1 and A2) present not A3)
=P( A 1 and A2 and not A3)/P(A1 and A2)
= (P(A1 and A2)- P(A1 int A2 int A3))/P(A1 and A2)
=(.06-.01)/.06
=.8333

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