Question

A firm manufactures PVC pipes in three plants viz, X, Y and Z. The daily production volumes from the three firms X, Y and Z are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s total production, if the selected pipe is a defective, then what is the probability that it was produced by plant Y?

Answer 1

Let A₁ be the daily volume of production by plant X

A₂ be the daily volume of production by plant Y

A₃ be the daily volume of production by plant Z.

Let B be the defective output we have to find P(B)

P(A₂/B) = `("P"("A"_2) * "P"("B"/"A"_2))/("P"("A"_1) * ("B"/"A"_1) + "P"("A"_2) * "P"("B"/"A"_2) + "P"("A"_3) * "P"("B"/"A"_3)`

= `(3/10 xx 0.04)/(1/5 xx 0.03 + 3/10 xx 0.04 + 1/2 xx 0.02)`

= `(0.12/10)/(0.03/5 + (3  xx  0.02)/5 + 0.02)`

= `(0.12/10)/((0.03 + 0.06 + 0.05)/5)`

= `0.12/10 xx 5/0.14`

= `0.12/0.28`

P(A₂/B) = `12/28`

= `3/7`

Probability that the defective pipe produced by plant Y = `3/7`