Question

A container holds 50 electronic components, of which 10 are defective. If 6 components are drawn at random from the container, the probability that at least 4 are not defective is . If 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is .

Answer 1

Let p be probability of container being non- defective

and q be probability of container being defective, so

p = 40/50 = 4/5 = 0.8

q = 10/50 = 1/5 = 0.2

Now we have

P(x) = ⁿCₓ  pˣ q ⁿ⁻ˣ

where n is no. of containers we draw at random, and P(x) is probability that x of them are non-defective.

 P(x ≥ 4) = P(4) + P(5) + P(6)

= ⁶C₄ (0.8)⁴ (0.2)² + ⁶C₅ (0.8)⁵ (0.2) + ⁶C₆ (0.8)⁶ (0.2)⁰

= 15 (0.4096) (0.04) + 6 (0.32768) (0.2) + 0.262144

= 0.24576 + 0.393216 + 0.262144

= 0.90112

Now,
Let q be probability of container being non- defective

and p be probability of container being defective, so we have

q = 40/50 = 4/5 = 0.8

p = 10/50 = 1/5 = 0.2

P(x) = ⁿCₓ  pˣ q ⁿ⁻ˣ

where n is no. of containers we draw at random, and P(x) is probability that x of them are defective.

P(x=3) = P(3) = ⁸C₃ (0.2)³ (0.8)⁵

= 56 (0.008)  (0.32768)

= 0.1468