Question
Download Solution PDFकिसी उद्योग में एक बीमारी की घटना इस प्रकार है कि श्रमिकों को इससे पीड़ित होने की 20% संभावना है। यादृच्छिक रूप से चुने गए 6 श्रमिकों में से, 4 या अधिक श्रमिकों के बीमारी से पीड़ित होने की प्रायिकता क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFव्याख्या:
दिया गया है
n = 6
⇒ p = P (बीमारी से पीड़ित) = 20/100 = 1/5
⇒ q = P (बीमारी से पीड़ित नहीं) = 1 - 1/5 = 4/5
⇒ p(x ≥ 4) = p(x = 4) + p(x = 5) + p(x = 6)
⇒ \(^6C_4(\frac{1}{5})^4 (\frac{4}{5})^2 +^6C_5 (\frac{1}{5})^5(\frac{4}{5})^1+^6C_6(\frac{1}{5})^6\)
= \(\frac{6!}{4!2!}\times\frac{16}{5^6}+ \frac{6!}{5!}\times\frac{4}{5^6}+ \frac{1}{5^6}\)
= \(\frac{265}{15625} = \frac{53}{3125}\)
∴ विकल्प (a) सही है
Last updated on May 30, 2025
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