A shipment of 2000 light bulbs contains 200 defective items

A shipment of 2,000 light bulbs contains 200 defective items and 1,800 good items. Five hundred bulbs are chosen at random and are tested, and the entire shipment is rejected if more than 25 bulbs from among those tested are found to be defective. What is the probability that the shipment will be accepted?

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

Picking a Defective bulb chance P = 200/2000 = 0.10

P( X < = 25) = P(X=25) + P(X=24) + P(X=23) + P(X=22) + P(X=21) + ....P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 500 25 ) * 0.1^25 * ( 1- 0.1 ) ^475 + ( 500 24 ) * 0.1^24 * ( 1- 0.1 ) ^476 + .....( 500 0 ) * 0.1^0 * ( 1- 0.1 ) ^500
= 0.0000354

P(The shipment will accept ) = 1 - P(Defect below or equal 25) = 1 - 0.0000354 = 0.9999646