There are three coins. One is a two-headed coin (having a head on both faces), another is a biased coin that comes up heads 75 \% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin?
There are three coins. One is a two-headed coin (having a head on both faces), another is a biased coin that comes up heads 75 \% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin?

Solution:

Let E_{1} represent the event of selecting a two-headed coin, E_{2} represent the event of selecting a biassed coin, and \mathrm{E}_{3} represent the event of selecting an unbiased coin. Let \mathrm{A} be the event in which the coin lands on its head.

Then P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=1 / 3

Because a headed coin has a head on both sides, it will display one.

Also P\left(A \mid E_{1}\right)=P (correct answer given that he knows) =1

And \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P} (coin shows head given that the coin is biased) =75 \%=75 / 100=3 / 4 And P\left(A \mid E_{3}\right)=P (coin shows head given that the coin is unbiased) =1 / 2

Given that the coin is showing head, the probability that it is two-headed is \mathrm{P} \left(\mathrm{E}{1} \mid \mathrm{A}\right).

We have used Bayes’ theorem to arrive at our conclusion.

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{3}\right)}

We get the result by swapping the values.

=\frac{\frac{1}{3} \cdot 1}{\frac{1}{3} \cdot 1+\frac{1}{3} \cdot \frac{3}{4}+\frac{1}{3} \cdot \frac{1}{2}}

=\frac{\frac{1}{3}}{\frac{1}{3}\left(1+\frac{3}{4}+\frac{1}{2}\right)}

=\frac{1}{\frac{9}{4}}=\frac{4}{9}

\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{4}{9}