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Bag I contains 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. then the probability that it was drawn from Bag I is

Solution

✅ Correct Option: 3

By Bayes' theorem, the required probability is 12⋅3712⋅37+12⋅511\dfrac{\frac{1}{2}\cdot\frac{3}{7}}{\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{5}{11}}.

The halves cancel, leaving 3/73/7+5/11=33/7768/77\dfrac{3/7}{3/7+5/11} = \dfrac{33/77}{68/77}.

Hence the probability is 3368\dfrac{33}{68}.

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