Question

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

There is some inforrmation that is missing in this question. It should read:

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 .

Answers
Part 1.   0.02
Part 2.  0.0375

Explanation
The probability is a chance of an event happening. It is calculated as;
probability = (Number of favourable outcome)/(Number of available outcome)

Part 1
6 are chosen at random. If 4 are not defective, then 2 are defective.
P(at least 4 are not defective) = 4/40 × 2/10
                                                = 1/10 ×1/5
                                                = 1/50
                                                 = 0.02

Part 2
8 are chosen at random. If 3 are defective, the 5 are not defective. 
P(3 are defective) = 3/40 × 5/10 
                             = 15/400
                             = 3/80
                            = 0.0375

Answer 2

Answer:

1.)0.91

2.)0.147

Step-by-step explanation:

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 (0.91)

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 (0.147)