Probability Practice Test 2

Question : 36 of 37

Marks: +1, -0

In a factory where toys are manufactured, machines A, B and C produce 25%, 35% and 40% of the total toys, respectively. Of their output, 5%, 4% and 2% respectively, are defective toys. If a toy drawn at random is found to be defective, what is the probability that it is manufactured by Machine B?

$17/69$
$28/69$
$35/69$
None of these

Validate

Solution:

Let A = Event of manufacturing on Machine A
B = Event of manufacturing on Machine B
C = Event of manufacturing on Machine C
D = Event of manufacturing on defective item
∴ P(A) = P(B) = P(C) =

1/3

P(D/A)=5/25,P(D/B)=4/35,P(D/C)=2/40

Now using Bayes’ theorem,

P(B/D)= {P(D/B). P(B)}/{P(D/A).P(A)+P(D/B).P(B)+P(D/C).P(C) }

={4/35×1/3}/{5/25×1/3+4/35×1/3+2/40×1/3}

=16/51

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