A container holds 50 electronic components, of which 10 are defective.

1. If 6 components are drawn at random from the container, the probability that at least 4 are not defective is?

A) 0.26

B) 0.42

C) 0.75

D) 0.91

E) 1.00

2. If 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is?

A) 0.147

B) 0.203

C) 0.300

D) 0.375

E) 0.750

1.) q = P (defective) = 10/50 = 0.2

p = P (not defective) = 1 - P (defective) = 1 - 0.2 = 0.8

P (x) = nCr p^x q^ (n-x)

P (x ≥ 4) = P (4) + P (5) + P (6) = 6C4 * (0.8) ^4 * (0.2) ^2 + 6C5 * (0.8) ^5 * 0.2 + (0.8) ^6 = 15 * 0.4096 * 0.04 + 6 * 0.32768 * 0.2 + 0.262144 = 0.24576 + 0.393216 + 0.262144 = 0.90112

Option D is the correct answer.

2.) p = P (defective) = 10/50 = 0.2

q = P (not defective) = 1 - P (defective) = 1 - 0.2 = 0.8

P (x) = nCr p^x q^ (n-x)

P (x = 3) = 8C3 * (0.2) ^3 * (0.8) ^5 = 8C3 * 0.008 * 0.32768 = 56 * 0.008 * 0.32768 = 0.1468

Option A is the correct answer.