Question

There are three coins. One is a two-headed coin another is a biased coin that comes up heads $75$% of the time and 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?

Answer 1

Let $E_1,E_2$, and $E_3$ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.
$\therefore P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

Let $A$ be the event that the coin shows heads.
A two-headed coin will always show heads.

$\therefore P\left(A|E_1\right)=P(coinshowingheads,giventhatitisatwo-headedcoin)=1$
Probability of heads coming up, given that it is a biased coin $=75$%

$\therefore P\left(A|E_2\right)=P(coinshowingheads,giventhatitisabiasedcoin)=\frac{75}{100}=\frac{3}{4}$

Since the third coin is unbiased, the probability that it shows heads is always $\frac{1}{2}$

$\therefore P\left(A|E_3\right)=P(coinshowingheads,giventhatitisanunbiasedcoin)=\frac{1}{2}$
The probability that the coin is two-headed, given that it shows heads, is given by $P\left(E_1|A\right)$.
By using Baye's theorem, we obtain

$P\left(E_1|A\right)=\frac{P\left(E_1\right)\cdot P\left(A|E_1\right)}{P\left(E_1\right)\cdot P\left(A|E_1\right)+P\left(E_2\right)\cdot P\left(A|E_2\right)+P\left(E_3\right)\cdot P\left(A|E_3\right)}$

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

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

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

$=\frac{4}{9}=0.44$