Question

A shipment of 1000 radios contains 15 defective ones. a) If an inspector selects 10 to sample, what is the % chance he will detect at least one defect? b) How many radios must be sampled to yield a 50+% chance of detecting a defect?

Answer 1

Given the large shipment size of radios, we can work out the probability of a defective one

P(A radio is defective) = 15/1000 = 0.015 => P(A radio is not defective) = 1 - 0.15 = 0.85

a) P(No defective radio in a 10-size sample) = (0.85)¹⁰    (different radios being defective or not are independent events)

Hence, P(At least one radio is defective) = 1 - (0.85)¹⁰ = 1 - 0.1969 = 0.8031

b) The number of trials to make until we see the first defect, say X, follows a geometric distribution, with probability p and distribution:

f(x) = (1-p)^x-1*p

No. of radios x to be sampled to sample for 50+% chance of detecting a defect, is given by the expected value of X

f(x) = (1-p)^x-1  = 0.5 => (1 - 0.15)^x-1 = 0.5

=>0.85^x-1 = 0.5

=> ( x-1 ) * log(0.85) = log (0.5)

=> (x - 1)*(-0.07058) = -0.301

=> x = 5.265

Hence, we need to sample about 6 or more radios for 50+% chance of detecting a defect.