tpes of defects. uat A, t. .2.3) denote the event that the system ha. . defeet o

Question

tpes of defects. uat A, t. .2.3) denote the event that the system ha. . defeet of typ" StOpes, that t-suloi abile r. tree, MAI U A2)-0.14 PKA UAj)-0.14 (a) Given that the systems has a type 1 defect, what is the prabability that it has a type 2 defect?(Round yeur ansuer to four deomal slaces Given that the system has type defect chet is the probablity that it has all three types of defects? (ound your siver to hour decimal places ) 00s Need Help? L s) " " h en-.mey-om, ..t is n-areb.blty that n ua nut b. dscovered,

Explanation / Answer

P(A1 and A2) = P(A1) + P(A2) - P(A1 or A2)
= 0.12 + 0.08 - 0.14 = 0.06

P(A1 and A3) = P(A1) + P(A3) - P(A1 or A3)
= 0.12 + 0.05 - 0.14 = 0.03

P(A2 and A3) = P(A2) + P(A3) - P(A2 or A3)
= 0.08 + 0.05 - 0.11 = 0.02

a)
P(A2 | A1) = P(A1 and A2)/P(A1) = 0.06/0.12 = 0.5

b)
P((A1 and A2 and A3) | A1) = P(A1 and A2 and A3)/P(A1) = 0.01/0.12 = 0.0833

c)
P(A1 or A2 or A3) = P(A1) + P(A2) + P(A3) - P(A1 and A2) - P(A1 and A3) - P(A2 and A3) + P(A1 and A2 and A3)
= 0.12 + 0.08 + 0.05 - 0.06 - 0.03 - 0.02 + 0.01
= 0.15

P(only A1) = P(A1) - P(A1 and A2) - P(A1 and A3) + P(A1 and A2 and A3)
= 0.12 - 0.06 - 0.03 + 0.01
= 0.04

P(only A2) = P(A2) - P(A1 and A2) - P(A2 and A3) + P(A1 and A2 and A3)
= 0.08 - 0.06 - 0.02 + 0.01
= 0.01

P(only A3) = P(A3) - P(A1 and A3) - P(A2 and A3) + P(A1 and A2 and A3)
= 0.05 - 0.03 - 0.02 + 0.01
= 0.01

Required probability = (P(only A1) + P(only A2) + P(only A3))/P(A1 or A2 or A3)
= (0.04 + 0.01 + 0.01)/0.15
= 0.4

d)
P(A3' | (A1 and A2)) = P(A1 and A2 and A3')/P(A1 and A2)

P(A1 and A2 and A3') = P(A1 and A2) - P(A1 and A2 and A3)
= 0.06 - 0.01
= 0.05

P(A3' | (A1 and A2)) = 0.05/0.06 = 0.8333

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