**Toolbox:**

- \(E_1=chosing\; a \;baised \;coin(two\;headed)\)
- \(E_2=chosing\; a \;un\;baised \;coin\)
- \(A\;getting\;a\;head\)
- \(there\;are\;'n'\;baised\;coin\;and\;(n+1)\;unbaised\;coin\;in\;a\;total\;of\;(2n+1)\;coin\)
- =\(P(E_1)P(a/E_1)+P(E_2)P(A/E_2)\)

\(P(E_1)=P(chosing\;a\;baised\;coin)\)

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

\(P(E_2)=P(chosing\;a\;unbaised\;coin)\)

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

\(P(A/E_1)=P(getting\;a\;head/it\;is\;baised\;(two\;headed\;coin))\)

=1

\(P(A/E_2)=P(getting\;a\;head/it\;is\;unbaised\;coin)\)

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

\(\Large\;P(A)=1\times\frac{n}{2n+1}+\frac{1}{2}\times\frac{(n+1)}{(2n+1)}=\frac{31}{42}\)[given]

=\(\Large\;\frac{2n+n+1}{2(2n+1)}=\frac{3n+1}{2(2n+1)}=\frac{31}{42}\)

=\(42(3n+1)=2\times31(2n+1)\)

=\(126n+42=124n+62\)

=\(2n=20\)

\(n=10\)