$\begin{array}{1 1} \frac{1}{3} \\ \frac{2}{3} \\ \frac{1}{2} \\ \frac{3}{4} \end{array} $

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- Let us define E\(_1\) Taking a fair coin for toss
- E\(_2\) Taking a baised coinfor toss
- P(taking a fair coin for toss/getting a head)
- P(E\(_1\)/A)\(\Large\frac{{p(E_1)}{P(A/E_1)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}\)

since taking any coin for toss in equally likely

P(\(E_1\))=p(\(E_2\))=\(\large\frac{1}{2}\)

P(A/\(E_1\))=P(getting head when fair coins is taken)

=\(\large\frac{1}{2}\)

P(A/\(E_2\))=P(getting a head when baiased coin is taken)

=\(1\)

=\(\Large\frac{\frac{1}{2}\times\frac{1}{2}}{\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times1}\)

=\(\Large\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}\)

=\(\Large\frac{1}{4}\times\)\(\frac{4}{3}\)=\(\frac{1}{3}\)

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