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

Question

A certain system can experience three different types of defects. Let A_j (i = 1, 2, 3) denote the event that the system has a defect of type j. Suppose that the following probabilities are true. P(A_1) = 0.12 P(A_2) = 0.07 P(A_3 = 0.05 P(A_1 U A_2) = 0.13 P(A_1 U A_3) = 0.14 P(A_2 U A_3) = 0.10 P(A_1 n A_2 n 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.) 0.5000 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 0.08333 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 0.0500 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.) 0.8333

Explanation / Answer

a. P(A2|A1) = P(A2 int A1)/P(A1)
= P(A2)+P(A1)-P(A2 union A2) / P(A1)
=.12+.07-.13 / .12
=.5

b.P(A1 int A2 int A3|A1)
= .01/.012
= 1/12

c.

c. P(A1 int A3)
= P(A1)+P(A3)-P(A1 union A3)
=.12+.05-.14
=.04

P( A1 int A2)
=P(A1) +P(A2)-P(A1 union A2)
=.12+.07-.13
=.06

P(A2 int A3)
=P(A2)+P(A3)-P(A2 union A3)
=.07+.05-.10
=.02

P( atleast 1 defect ) = P(A1 union A2 union A3)
=.12+.07+.05 - .04-.06-.02+.01
=.13

So,P(Given atleast 1 defect, exactly 1 defect ) = ?

= P(exactly 1 defect)/.13

= P(A1) - P(A1 int A2) - P(A1 int A3) +P(A1 int A2 int A2) + P(A2) - P(A1 int A2) - P(A2 int A3) +P(A1 int A2 int A3) + P(A3) - P(A3 int A2) - P(A3 int A1) +P(A1 int A2 int A2) / .13

= (.12+.07+.05 -24+3(.01))/.13

=.231

d.

P(A3'|A1 int A2)

= P(A1 int A2) - P(A1 int A2 int A2)/.5

= .06-.01/.6

=5/6

= .833

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