Question
Download Solution PDFयदि P(A) = 0.5, P(B) = 0.7 और P(A ∩ B) = 0.3 हैं, तो P(A' ∩ B') + P(A' ∩ B) + P(A ∩ B') का मान क्या है ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
- P(A') = 1 - P(A)
- P(A' ∩ B') = P((A ∪ B)') {A और B के नहीं होने की प्रायिकता}
- P(A और B नहीं) = P(A ∩ B') = P(A) - P(A ∩ B)
- P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
गणना:
दिया गया है: P(A) = 0.5, P(B) = 0.7 और P(A ∩ B) = 0.3,
⇒ P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
⇒ P(A ∪ B) = 0.5 + 0.7 - 0.3 = 0.9__(i)
अब A और B के नहीं होने की प्रायिकता,
P(A' ∩ B') = P((A ∪ B)') = 1 - P(A ∪ B)
⇒ P(A' ∩ B') = 1 - 0.9 {(i) से }
⇒ P(A' ∩ B') = 0.1 __(ii)
B के होने और A के नहीं होने की प्रायिकता,
P(A' ∩ B) = P(B) - P(A ∩ B)
⇒ P(A' ∩ B) = 0.7 - 0.3 = 0.4 __(iii)
A के होने और B के नहीं होने की प्रायिकता,
P(A ∩ B') = P(A) - P(A ∩ B)
⇒ P(A' ∩ B) = 0.5 - 0.3 = 0.2 __(iv)
(ii), (iii) और (iv) से,
P(A' ∩ B') + P(A' ∩ B) + P(A ∩ B') = 0.1 + 0.4 + 0.2 = 0.7
∴ सही विकल्प (2) है।
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