Question

There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads $75$% of the times and third is also a biased coin that comes up tails $40$% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?

Answer 1

If there are $3$ coins.
Let these are $A,B,C$ respectively.
For coin A $\to$ Probability of getting head $P\left(H/A\right)=1$

For coin B $\to$ Probability of getting head $P\left(H/B\right)=\frac{3}{4}$

For coin C $\to$ Probability of getting head $P\left(H/C\right)=0.6$
we have to find $P\left(\frac{A}{H}\right)$ = Probability of getting H by coin A.
So, we can use formula
$P\left(\frac{A}{H}\right)=\frac{P\left(\frac{H}{A}\right).P\left(A\right)}{P\left(\frac{H}{A}\right).P\left(A\right)+P\left(\frac{H}{B}\right).P\left(B\right)+P\left(\frac{H}{C}\right).P\left(C\right)}$

Here, $P\left(\frac{H}{A}\right)=1,P\left(\frac{H}{B}\right)=\frac{3}{4},P\left(\frac{H}{C}\right)=0.6$

Put value in formula so, $P\left(\frac{A}{H}\right)=\frac{1.\frac{1}{3}}{1.\frac{1}{3}+\frac{3}{4}.\frac{1}{3}+\mathrm{0.6.}\frac{1}{3}}$

$=\frac{1}{1+0.75+0.6}$

$=\frac{100}{235}$

$=\frac{20}{47}$