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jacjAleshaw
Answered

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 ?
A0.26
B0.42
C0.75
D0.91
E1.00
  1. If 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is ?
A0.147
B0.203
C0.300
D0.375
E0.750

Verified Answer :

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