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Bag I contains 3 red and 4 black balls while another 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. The probability that it was drawn from Bag II is :

Solution

✅ Correct Option: 1

By Bayes theorem:

P(II∣R)=P(II)P(R∣II)P(I)P(R∣I)+P(II)P(R∣II)P(II|R) = \frac{P(II) P(R|II)}{P(I) P(R|I) + P(II) P(R|II)}

=12⋅51112⋅37+12⋅511= \frac{\frac{1}{2} \cdot \frac{5}{11}}{\frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{5}{11}}

=522314+522= \frac{\frac{5}{22}}{\frac{3}{14} + \frac{5}{22}}

LCD = 154: =35/15433/154+35/154=3568= \frac{35/154}{33/154 + 35/154} = \frac{35}{68}.

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