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

2. angles qrt and qst both measures 156 degrees.

step-by-step explanation:  

we have been given a diagram of a circle and we are told that measure of arc qvt is 156 degrees.  

we can see from our given diagram that angle qrt and angle qst subtends to arc qvt.

since we know that angles subtended by the same arc at the circumference are always equal.

by angle subtended by same arc theorem, measure of angle qrt and angle qst will be equal to the measure of arc qvt.

therefore, angles qrt and qst both measures 156 degrees and 2nd statement is the correct choice.