Q9 Three machines P1, P2, and P, in a
3
certain factory produce 50%, 50% and
25%, respectively, of the total daily output
of electric tubes. It is known that 5% of
the tubes produced by each of the
machines P, and P2 are defective and
that 4% of those produced on P, are
defective. If one tube is picked up at
random from a day’s production,
calculate the probability that the
defective tube was produced on machine
P1
Given : Three machines, P1, P2, P3 in a factory produce 50%, 25% and 25% respectively in a day
To find : probability that the defective tube was produced on machine
P1
probability of defective item
Solution:
Let say Total Production by Machine = 1000
Production by Machine P1 = (50/100) 1000 = 500
Production by Machine P2 = (25/100) 1000 = 250
Production by Machine P3 = (25/100) 1000 = 250
Defective output by P1 = (5/100) 500 = 25
Defective output by P2 = (5/100) 250 = 12.5
Defective output by P3 = (4/100) 250 = 10
Total Defectives = 25 + 12.5 + 10 = 47.5
probability that the defective tube was produced on machine P1 = 25/47.5
= 0.5263
= 52.63 %
Probability that items was defective = 47.5/1000
= 0.0475
= 4.75 %
0.0475 (4.75 % ) is the probability that item was defective.
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