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Question

# A bag contains 1 white and 6 red balls, and a second bag contains 4 white and 3 red balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be white in colour. Find the probability that the drawn ball was from the first bag.

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Solution

## Let A, E1 and E2 denote the events that the ball is white, bag I is chosen and bag II is chosen, respectively. $\therefore P\left({E}_{1}\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}P\left({E}_{2}\right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}P\left(A/{E}_{1}\right)=\frac{1}{7}\phantom{\rule{0ex}{0ex}}P\left(A/{E}_{2}\right)=\frac{4}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\mathrm{Required}\mathrm{probability}=P\left({E}_{1}/A\right)=\frac{P\left({E}_{1}\right)P\left(A/{E}_{1}\right)}{P\left({E}_{1}\right)P\left(A/{E}_{1}\right)+P\left({E}_{2}\right)P\left(A/{E}_{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1}{2}×\frac{1}{7}}{\frac{1}{2}×\frac{1}{7}+\frac{1}{2}×\frac{4}{7}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+4}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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