Question

The department of quality control of a factory has establish a lot that contains 90% or more of good items is a lot of good quality. It has been establish to the following plan. A sample is taken of 4 items, if 1 o none is defective then the lot is accepted. A second sample is taken with 2 items, if the second sample does not contain defective items then the lot is accepted. If not, then it will be not accepted. Supposed the lot is very big and that the probability of extracting an item that is defective is constant. (Recommendation: Use the Tree Diagram to visualize the process of acceptance and non-accpetance)

a) Calculate the probability of accepting a lot that contains 25% of Bad Items.

b)Calculate the probability of not accepting a lot that contains 5% of Bad Items.

b)

Answer 1

Ans a) P(Item is bad) = 0.25 => P(Item is not bad) = 0.75

Now, P(acceptance of first sample)

= probability of getting 0 or 1 defective items in first sample

=  0.75⁴ + 0.75³ * 0.25 = 0.3164 + 0.1055 = 0.422

Now, P(acceptance of second sample)

= Probability of getting all non-defective items in second sample

= 0.75² = 0.5625

Hence, as the 2 sampling events are independent,

P(Acceptance of the lot) = P(Both samples pass the test) = 0.422*0.5625 = 0.237

Ans b) Similar to above, using P(Item is bad) = 0.05, we get

P(Acceptance of first sample) = 0.95⁴ + 0.95³ * 0.05 = 0.8145 + 0.0429 = 0.991

P(Acceptance of second sample) = 0.95² = 0.9025

Hence, as the 2 sampling events are independent,

P(Acceptance of the lot) = P(Both samples pass the test) = 0.991*0.9025 = 0.8944