There are three coins. One is two-headed coin (having head on bot… - Vidyadip

Question

There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tail 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?

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Answer

A be the event of choosing two - headed coin,
B be the event of choosing a biased coin that comes up head 75% of the times,
C be the event of choosing a biased coin that comes up tail 40% of the times and
E be the event of getting a head.
Now,
$\text{P(A)}=\text{P(B)}=\text{P(C)}=\frac{1}{3}$ and
$\text{P}(\text{E}|\text{A})=1,\text{P}(\text{E}|\text{B})=75\%=\frac{75}{100}=\frac{3}{4}$ and $\text{P}(\text{E}|\text{C})=60\%=\frac{60}{100}=\frac{3}{5}$
So, using Bayes' theorem, we get
P (the head shown was of two - headed coin) = P(A|E)
$=\frac{\text{P(A)}\times\text{P}(\text{E}|\text{A})}{\text{P(A)}\times\text{P}(\text{E}|\text{A})+\text{P(B)}\times(\text{E}|\text{B})+\text{P(C)}\times\text{P}(\text{E}|\text{C})}$
$=\frac{\Big(\frac{1}{3}\times1\Big)}{\Big(\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{3}{5}\Big)}$
$=\frac{\Big(\frac{1}{3}\Big)}{\Big(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\Big)}$
$=\frac{\Big(\frac{1}{3}\Big)}{\Big(\frac{20+15+12}{60}\Big)}$
$=\frac{\Big(\frac{4}{3}\Big)}{\Big(\frac{47}{60}\Big)}$
$=\frac{60}{3\times47}$
$=\frac{20}{47}$
So, the probability that the head shown was of a two-headed coin is $=\frac{20}{47}$.
Disclaimer: The answer given in the book is incorrect. The same has been corrected here.

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