The occurrence of a disease in an industry is such that the workers have 20% chance of suffering from it. What is the probability that out of 6 workers chosen at random, 4 or more will suffer from the disease?

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NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024)
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  1. 53/3125
  2. 63/3125
  3. 73/3125
  4. 83/3125

Answer (Detailed Solution Below)

Option 1 : 53/3125
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Detailed Solution

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Explanation:

Given

n = 6

⇒ p = P (suffering from disease) = 20/100 = 1/5

⇒ q = P (not suffering from disease) = 1 - 1/5 = 4/5

⇒ p(x ≥ 4) = p(x = 4) + p(x = 5) + p(x = 6)

⇒ \(^6C_4(\frac{1}{5})^4 (\frac{4}{5})^2 +^6C_5 (\frac{1}{5})^5(\frac{4}{5})^1+^6C_6(\frac{1}{5})^6\)

\(\frac{6!}{4!2!}\times\frac{16}{5^6}+ \frac{6!}{5!}\times\frac{4}{5^6}+ \frac{1}{5^6}\) 

\(\frac{265}{15625} = \frac{53}{3125}\)

∴ Option (a) is correct

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