Question
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Detailed Solution
Download Solution PDFदिया गया है:
केतली A में 3 नीली और 4 हरी गेंदें हैं
केतली B में 5 नीली और 6 हरी गेंदें होती हैं
एक केतली से यादृच्छया निकाली गई एक गेंद नीली थी
संकल्पना:
बेय की प्रमेय:
हल:
मान लीजिए कि निकाली गई गेंद के नीले होने की घटना B है और E1 यह घटना है कि गेंद केतली 1 से निकाली गई है और E2 घटना है कि गेंद केतली 2 से निकाली गई है।
∴ P(B) = P(B ∩ E1) + P(B ∩ E2)
⇒ P(B) = \(\frac{1}{2}\times\frac{3}{7} + \frac{1}{2}\times\frac{5}{11} \)
= 34/77
∴ P(E2/B) = \(\frac{P(E_1 \cap B)}{P(B)} = \frac{P(E_2) P(B/E_2))}{P(B)}\)
= \(\frac{\frac{1}{2}\times \frac{5}{11}}{\frac{34}{77}}\)
= 35/68
Last updated on Jun 19, 2025
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