Answer:
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
