Let’s consider

\[{{E}_{1}}\] = Event that the coin is fair

\[{{E}_{2}}\] = Event that the coin is \[2\] headed

And H = Event that the tossed coin gets head.

Now,

\[P({{E}_{1}})\text{ }=\text{ 1/2},\text{ }P({{E}_{2}})\text{ }=\text{ 1/2},\text{ }P(H/{{E}_{1}})\text{ }=\text{ 1/2},\text{ }P(H/{{E}_{2}})\text{ }=\text{ }1\]

Using Baye’s Theorem, we get

Therefore, the required probability is \[1/3\].