A factory has three machines that all manufacture the same engine part. From long historical records the company has determined that the first machine produces 4% defectives, the second machine produces 6% defectives, and the third machine produces 1% defectives. Suppose also that the first machine produces 30%, the second machine 20%, and the third machine 50% of the engine parts. An engine part is selected at random. If the part is defective, what is the probability that was produced by the first machine?

Step-by-step explanation:

P(D/M₁) = .04 ( probability of defect from a 1 st machine )

P(D/M₂) = .06

P(D/M₃) = .01

P( M₁ )= .3 ( probability of manufacture from 1 st machine )

P(M₂) = .2

P(M₃)= .5

P( M₁/D) ( Probability of manufacture from 1 st machine , given it is defective )

= P( M₁ )xP(D/M₁) / [P( M₁ )xP(D/M₁) +P( M₂ )xP(D/M₂)+P( M₃ )xP(D/M₃)]

= .3 x .06 / .3 x .06 + .2 x .06 + .5 x .01

= .018 / .035

= 18/35

took the test it's 94

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.