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Bag A contains 3 red and 4 black balls, and Bag B contains 5 red and 'n' black balls. One ball is drawn at random from one of the bags and it is found to be red. If the probability that it was from bag B is 35/68, then the value of 'n' is:

Solution

✅ Correct Option: 4

By Bayes' theorem with equal bag probabilities, 55+n37+55+n=3568\dfrac{\frac{5}{5+n}}{\frac37+\frac{5}{5+n}}=\dfrac{35}{68}.

Put u=55+nu=\dfrac{5}{5+n}: 68u=15+35u68u=15+35u, so u=511u=\dfrac{5}{11}.

Thus 5+n=115+n=11 and n=6n=6.

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