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Urn A contains 1 white, 2 black and 3 red balls, Urn B contains 2 white, 1 black and 1 red ball and Urn C contains 4 white, 5 black and 3 red balls. One Urn is chosen at random and two balls are drawn. If it is found that there is one white and one red, Then the probability that they come from A is:

Solution

✅ Correct Option: 4

P(E∣A)=1×3(62)=315=15P(E|A)=\frac{1\times3}{\binom{6}{2}}=\frac{3}{15}=\frac15, P(E∣B)=2×1(42)=13P(E|B)=\frac{2\times1}{\binom{4}{2}}=\frac13, P(E∣C)=4×3(122)=1266=211P(E|C)=\frac{4\times3}{\binom{12}{2}}=\frac{12}{66}=\frac{2}{11}.

Each urn has prior 13\frac13, so by Bayes' theorem the priors cancel:

P(A∣E)=1/51/5+1/3+2/11=33/165118/165=33118P(A|E)=\frac{1/5}{1/5+1/3+2/11}=\frac{33/165}{118/165}=\frac{33}{118}.

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