A car manufacturer who is producing cars at the rate of 4000 cars a month requires to procure ignition equipment from vendors. The company has a policy of placing orders for 12000 ignition equipment every time they order. The vendor has indicated that 3 % of the ignition equipment is likely to be faulty in each shipment. Since it is not possible to test every ignition equipment by the car manufacturer, the manufacturer decides to randomly sample 200 ignition equipment and if more than 2 of them are found to be defective the entire batch of order shall be returned back.

a) What sampling method would you suggest for the car manufacturer?
b) What is the probability that the batch will be returned back to the vendor.

a) ok so first you have to suggest a sampling method for the car manufacturer, since its given that he randomly decides to test 200 ignition equipment then best strategy is simple random sampling (all subsets of frame are given equal probability).. review sampling methods to understand them better.

b) probability that batch will be returned to vendor: no guessing here, simply refer to the question (batch will be returned if more than 2 of them are found to be defective). this is the thing we have to calculate,

P(X>2) = 1 - P(X=0) - P(X=1) - P(X=2)... this should be obvious, the only other way is that you sum all possibilities of X from 3 to 200!

Binomial distribution with n = 200 and p = 0.03

P(X=0) = 200C0 * (0.03)^0 * (0.97)^200 = 0.00226124
P(X=1) = 200C1 * (0.03)^1 * (0.97)^199 = 0.013987
P(X=2) = 200C2 * (0.03)^2 * (0.97)^198 = 0.04304

P(X>2) = 1 - 0.00226124 - 0.013987 - 0.04304 = 0.9407