Let, A1 be the event that the drawn bolt is manufactured by machine A
A2 be the event that the drawn bolt is manufactured by machine B
A3 be the event that the drawn bolt is manufactured by machine C
B be the event that the drawn bolt is defective
Now, we are given,
P(A1) = 25% = 0.25, P(A2) = 30% = 0.3, P(A3) = 45% = 0.45, P(B | A1) = 7% = 0.07, P(B | A2) = 6% = 0.06, P(B | A3) = 4% = 0.04
The required probability
= The probability that the defective bolt was manufactured by machine C
= The probability that the bolt was manufactured by machine C given it was defective
= P(A3 | B)
Here the events A1, A2 and A3 are mutually exclusive and exhaustive.
By using Bayes' theorem,
P(A3 | B) = "\\frac{P(A_3).P(B |A_3)}{P(A_1).P(B |A_1)+P(A_2).P(B |A_2)+P(A_3).P(B |A_3)}"
= "\\frac{0.04 \\times 0.45}{0.07 \\times 0.25+0.06 \\times 0.3+0.04 \\times 0.45}"
= 0.336 (rounded to 3 decimal places)
Answer: If a bolt drawn at random from the production is found to be defective, the probability that it was manufactured by machine C is 0.336.
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