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Question

The content of 3 urns 1,2,3 are as follows : 1 white, 2 black, 3 red balls; 2 white, 1 black, 1 red balls; 4 white, 5 black, 3 red balls. One urn is selected at random and two balls are drawn and these comes out to be white and red. Then the probability that they come from

A
2nd urn is 55118
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B
3rd urn is 30118
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C
1st urn is 33118
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D
3rd urn is 59118
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Solution

The correct option is C 1st urn is 33118
Let Ei be the event of choosing urn 'i', (i=1,2,3). Let E be the event of getting one white and one red ball when 2 balls are drawn from the chosen urn.
P(E1)=13,P(E2)=13 and P(E3)=13
P(EE1)=1×36C2, P(EE2)=2×14C2 and P(EE3)=4×312C2
Using total probability theorem, we have:
P(E)=P(Ei).P(E/Ei)=13[1×3(6C2)+2×1(4C2)+4×3(12C2)]=13[15+13+211]
Using Baye's therorem
P(E1/E)=13(15)13(15+13+211)=3333+55+30=33118
P(E2/E)=13(13)13(15+13+211)=55118
P(E3/E)=13(211)13(15+13+211)=30118

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