Question

A component is purchased from 3 suppliers, A, B, and C, where the suppliers have

respective defective rates of 2%, 6%, and 4%. Of all the components purchased, 20%

comes from supplier A, 50% from supplier B, and 30% from supplier C, that is, each

shipment comes from each of these suppliers with these probabilities. The company uses

the following quality control policy. A sample of 15 units is randomly selected from each

shipment of components. If at most 2 defective units are found in the sample, then the

entire shipment is accepted; otherwise, the entire shipment is rejected.

Determine the following:

(a-1) probability that a shipment will be accepted given that it came from Supplier A.

(a-2) probability that a shipment will be accepted given that it came from Supplier B.

(a-3) probability that a shipment will be accepted given that it came from Supplier C.

(a-4) Using the law of total probability and your responses to parts (a-1)-(a-3) above,

determine the probability that a shipment will be accepted.

(b-1) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier A.

(b-2) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier B.

(b-3) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier C.

(b-4) Using the law of total expectation and your responses to parts (b-1)-(b-3) above,

determine the expected number of defectives in a random sample of 15 units.

Answer 1

(a-1) probability that a shipment will be accepted given that it came from Supplier A.

Let S be the event that the shipment is accepted and R be the event that the shipment is rejected.

Let A, B, C be the event that the shipment comes from the suppliers A, B and C respectively.

Let X be the number of defective units found in the sample of 15.

P(S | A) = Probability that at most 2 defective units are found in the sample of 15 where the probability of defect is 2% (0.02)

= P(X = 0) + P(X = 1) + P(X = 2)

= ¹⁵C₀ * 0.02⁰ * 0.98¹⁵ +  ¹⁵C₁ * 0.02¹ * 0.98¹⁴ +  ¹⁵C₂ * 0.02² * 0.98¹³

= 0.9970

(a-2) probability that a shipment will be accepted given that it came from Supplier B.

P(S | B) = Probability that at most 2 defective units are found in the sample of 15 where the probability of defect is 6% (0.06)

= P(X = 0) + P(X = 1) + P(X = 2)

= ¹⁵C₀ * 0.06⁰ * 0.94¹⁵ +  ¹⁵C₁ * 0.06¹ * 0.94¹⁴ +  ¹⁵C₂ * 0.06² * 0.94¹³

= 0.9429

(a-3) probability that a shipment will be accepted given that it came from Supplier C.

P(S | C) = Probability that at most 2 defective units are found in the sample of 15 where the probability of defect is 4% (0.04)

= P(X = 0) + P(X = 1) + P(X = 2)

= ¹⁵C₀ * 0.04⁰ * 0.96¹⁵ +  ¹⁵C₁ * 0.04¹ * 0.96¹⁴ +  ¹⁵C₂ * 0.04² * 0.96¹³

= 0.9797

(a-4) Using the law of total probability and your responses to parts (a-1)-(a-3) above,

determine the probability that a shipment will be accepted.

P(S) = P(A) P(S | A) + P(B) P(S | B) + P(C) P(S | C)

= 0.2 * 0.9970 + 0.5 * 0.9429 + 0.3 * 0.9797

= 0.9648

(b-1) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier A.

By binomial distribution, E(X | A) = np

= 15 * 0.02 = 0.3

(b-2) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier B.

E(X | B) = 15 * 0.06 = 0.9

(b-3) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier C.

E(X | C) = 15 * 0.04 = 0.6

(b-4) Using the law of total expectation and your responses to parts (b-1)-(b-3) above,

determine the expected number of defectives in a random sample of 15 units.

Expected number of defectives in a random sample of 15 units is,

E(X) = E(X | A) * P(A) + E(X | B) * P(B) + E(X | C) * P(C)

= 0.3 * 0.2 + 0.9 * 0.5 + 0.6 * 0.3

= 0.69