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An urn contains two type A coins and one type B coin.When a type A coin is flipped, it comes up heads with probability1/4,where as when a type B coin is flipped,it comes up heads with probability 3/4. A coin is randomly chosen from the urn and flipped. Given that the flip landed on heads, what is the probability that it was a type A coin?

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Probability of getting head = (2/3)*(1/4)+(1/3)*(3/4)

probability of getting head from type A coin = 2/3 * 1/4 

so desired conditional probability = (2/3)*(1/4) 
                                                  --------------------------------  =2/5
                                                  (2/3)*(1/4)+(1/3)*(3/4)

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It is a question  of Bayes' Theorem.

Let we have to find P(type A coin |  flip landed on heads) which is given by :

P(type A| head )            =        ( P(type A) * P(head | type A)) / (P(type A) * P(head | type A)) + (P(type B) * P(head | type B))

Since 2  type A coin and 1 type B coin

P(type A)      =    2/3                and                  P(type B)  =    1/3

Given P(head | type A)   =   1/4          and         P(head | type B)   =   3/4

Substituting in Bayes' Theorem formula , we have :

P(type A | head )         =     (2/3 * 1/4 ) / ( 2/3 * 1/4 + 1/3 * 3/4)

                                 =      (1/6) /  ( 1/6 + 1/4)

                                 = 2/5           =       0.4

Such questions are preferrable to be solved by using tree diagram.

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