Question 1197263
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three biased coins C1, C2, C3 {{{highlight(cross(ie))}}} <U>lie</U> on the table.
Their probabilities of falling heads are 1/3, 2/3 and 1, respectively. 
A coin is picked at random, tossed twice, and observed to fall heads both times.
Calculate the probability that it is C1.
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<pre>
They want you calculate the conditional probability that the randomly picked coin is C1 
given that tossed twice it produces (Head,Head).


    P(tossing one of the three coins twice gives HH) = {{{(1/3)*((1/3)^2 + (2/3)^2 + 1^2)}}} = {{{(1/3)*(1/9 + 4/9 + 9/9)}}} = 

                                                     = {{{(1/3)*(14/9)}}} = {{{14/27}}}.


    P(choosing C1 and (tossing C1 twice gives HH)) = {{{(1/3)*(1/3)^2}}} = {{{1/27}}}.


Therefore, the <U>ANSWER</U> is

    P = P(choosing C1 and (tossing C1 twice gives HH)) / P(tossing one of the three coins twice gives HH) = {{{((1/27))/((14/27))}}} = {{{27/(27*14)}}} = {{{1/14}}}.
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Solved, with full explanations.