Assume that both inspectors inspect every item and that if an item has no flaw, then neither inspector will detect a flaw. a. Assume that the probability that an item has a flaw is 0.10. If an item is passed by the first inspector, what is the probability that it actually has a flaw?
b. Assume that the probability that an item has a flaw is 0.10. If an item is passed by both inspectors, what is the probability that it actually has a flaw?

a) 0.011

b) 0.0032

Step-by-step explanation:

first lets say A₁ denotes the event that first inspector detects a flaw and A₂ is event that the law will be detected by the second inspector,

also let B denote the event that an item has a flaw.

a)

Given that;

P(B) = 0.1

so

P(A₁ / B ) = 0.9,  P(A₂ / B ) = 0.7

using Baye's rule, the probability that an item has a flaw if it was passed by the first inspector is;

P(B / A₁ⁿ ) =  [P(A₁ⁿ / B ) P(B)] / [ P(A₁ⁿ / B ) P(B) + P(A₁ⁿ / Bⁿ ) P(Bⁿ) ]

the probability that the first inspector will not detect the flaw if it actually exist is;

P(A₁ⁿ / B ) = 1 - P(A₁ / B )

= 1 - 0.9

= 0.1

Since there are no false detections, the probability that the first inspector will not detect the flaw, given that item does not have flaws is;

P(A₁ⁿ / Bⁿ ) = 1.0

so the probability that the item has flaw if it was passed by the first inspector is;

P(B / A₁ⁿ ) =  [(0.1) (0.1)] / [ (0.1 ) (0.1) + (1.0) (1 - 0.1) ]

= 0.01 / ( 0.01 + 0.9)

= 0.01 / 0.91

= 0.011

b)

Also by Bayes rule, the probability that the item has a flaw if it was passed by both inspectors is;

P(B / A₁ⁿ ∩ A₂ⁿ) =  [ P(A₁ⁿ ∩ A₂ⁿ / B) P(B)] / [ P(A₁ⁿ ∩ A₂ⁿ / B) P(B) + P(A₁ⁿ ∩ A₂ⁿ / Bⁿ) P(Bⁿ)]

Now since inspectors function independently, the probability that both inspectors will not detect the flaw if it actually exists is;

P(A₁ⁿ ∩ A₂ⁿ / B) = P(A₁ⁿ / B) P(A₂ⁿ / B)

= ( 1 - 0.9 ) ( 1 - 0.7)

= 0.1 * 0.3

= 0.03

since there are no false detections, the probability that both inspectors will not detect the flaw, given that item does not have a flaw is;

P(A₁ⁿ ∩ A₂ⁿ / Bⁿ) = 1.0

therefore

the probability that the item has a flaw if it was passed by both inspectors is;

P(B / A₁ⁿ ∩ A₂ⁿ) = [(0.03)(0.1)] / [(0.03)(0.1) + (1.0)(1-0.1)

= 0.003 / 0.93

= 0.0032

no idea

step-by-step explanation: