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Bag I contains 4 red and 3 black balls and bag II contains 6 red and 5 black balls. One ball is drawn at random from one of the bags, the probability that it is a red ball, is

Solution

✅ Correct Option: 1

Each bag is equally likely, so P(red)=12(47)+12(611)P(\text{red}) = \frac12\left(\frac47\right) + \frac12\left(\frac{6}{11}\right).

This equals 27+311=22+2177=4377\frac{2}{7} + \frac{3}{11} = \frac{22+21}{77} = \frac{43}{77}.

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