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Urn I contains 6 red balls and 4 black balls and Urn II contains 4 red balls and 6 black balls. One ball is drawn at random from Urn I and placed in Urn II. If one ball is drawn at random from Urn II, then the probability that it is a red ball is :

Solution

✅ Correct Option: 3

Case 1: Red drawn from Urn I (prob 6/106/10), then Urn II has 5R, 6B; P(red)=5/11P(\text{red})=5/11.

Case 2: Black drawn (prob 4/104/10), Urn II has 4R, 7B; P(red)=4/11P(\text{red})=4/11.

Total =610⋅511+410⋅411=46110=2355=\frac{6}{10}\cdot\frac{5}{11}+\frac{4}{10}\cdot\frac{4}{11}=\frac{46}{110}=\frac{23}{55}.

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