A factory produces bulbs.

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)