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)
=\(1\)
=\(\Large\frac{\frac{1}{2}\times\frac{1}{2}}{\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times1}\)
=\(\Large\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}\)
=\(\Large\frac{1}{4}\times\)\(\frac{4}{3}\)=\(\frac{1}{3}\)