Question

A company has three machines B1, B2, and B3, for making resistors. It has been observed:

Machine B1 produces 80% of resistors within 50 Ω of the nominal value

Machine B2 produces 90% of resistors within 50 Ω of the nominal value

Machine B3 produces 60% of resistors within 50 Ω of the nominal value

Each hour, machine B1 produces 3000 resistors
B2 produces 4000 resistors
B3 produces 3000 resistors

All of the resistors are mixed together at random in one bin and packed for shipment.

What is the probability that the company ships a resistor that is within 50 Ω of the nominal value?

Answer 1

Total resistors = 3000 + 4000 + 3000 = 10K

P(B1) = 3/10 = 0.3
P(B2) = 4/10 = 0.4
P(B3) = 3/10 = 0.3

Also, we are given here that:
P(within 50 of nominal value | B1) = 0.8
P(within 50 of nominal value | B2) = 0.9
P(within 50 of nominal value | B3) = 0.6

Using law of total probability, we get here:
P(within 50 of nominal value) = P(within 50 of nominal value | B1) P(B1) + P(within 50 of nominal value | B2) P(B2) + P(within 50 of nominal value | B3) P(B3)

P(within 50 of nominal value) = 0.8*0.3 + 0.9*0.4 + 0.6*0.3 = 0.78

Therefore 0.78 is the required probability here.