Question
Download Solution PDFUrn A consists 3 blue and 4 green balls while another urn B consists 5 blue and 6 green balls. One ball is drawn at random from one of the urns and it is found to be blue. Determine the probability that it was drawn from urn B?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Urn A consists 3 blue and 4 green balls
Urn B consists 5 blue and 6 green balls
One ball drawn at random from one of the urns was blue
Concept:
Baye's Theorem :
Solution:
Let B be the event that the ball drawn is blue and E1 be the event that the ball is drawn from urn 1 and E2 be the event that the ball is drawn from urn 2.
∴ 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 May 26, 2025
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