Question
Download Solution PDFThe 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?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
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
Last updated on May 30, 2025
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