Question:
A factory produces bulbs. The probability that any one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability that
(i) none of the bulbs is defective
(ii) exactly two bulbs are defective
(iii) more than 8 bulbs work properly
Solution:
Let’s assume X to be the random variable denoting a bulb to be defective.
Here, n = 10, p = 1/50, q = 1 – 1/50 = 49/50
We know that, P(X = r) = nCr pr qn – r
(i) None of the bulbs is defective, i.e., r = 0
P(x = 0) = 10C0 (1/50)0(49/50)10 – 0 = (49/50)10
(ii) Exactly two bulbs are defective
So, P(x = 2) = 10C2 (1/50)2(49/50)10 – 2
= 45.498/5010 = 45 x (1/50)10 x 498
(iii) More than 8 bulbs work properly
We can say that less than 2 bulbs are defective
P(x < 2) = P(x = 0) + P(x = 1)
