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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

Respuesta :

nobillionaireNobley
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.

Answer:

0.91

0.147

Step-by-step explanation:

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

0.91

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

0.147

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