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, find the probability that the selected pipe is a defective one

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).

Clearly, A₁, A_2, and A₃ are mutually exclusive and exhaustive events.

P(A₁) = `2000/10000 = 1/5`

P(A₂) = `3000/10000 = 3/5`

P(A₃) = `5000/10000 = 1/2`

P(B/A₁) = `3/100` = 0.03

P(B/A₂) = `4/100` = 0.04

P(B/A₃) = `2/100` = 0.02

P(B) = P(A₁) . P(B/A₁)  + P(A₂) . P(B/A₂) + P(A₃) . P(B/A₃)

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

= `0.03/5 + (3 xx 0.02)/5 + 0.01`

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

= `0.14/5`

P(B) = `14/500`

= `7/250`

Probability that the selected pipe is a defective one = `7/250`