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?     

Answer 1

\[\text{Let}: \]
\[\text{ A be the event of choosing two - headed coin, }  \]
\[\text{ B be the event of choosing a biased coin that comes up head 75 % of the times, }  \]
\[\text{ C be the event of choosing a biased coin that comes up tail 40 % of the times and } \]
\[\text{ E be the event of getting a head . } \]
\[\text{ Now } , \]
\[P\left( A \right) = P\left( B \right) = P\left( C \right) = \frac{1}{3} \text{ and} \]
\[P\left( E|A \right) = 1, P\left( E|B \right) = 75 % = \frac{75}{100} = \frac{3}{4} and P\left( E|C \right) = 60 % = \frac{60}{100} = \frac{3}{5}\]
\[\text{ So, using Bayes' theorem, we get } \]
\[P\left( \text{ the head shown was of two - headed coin } \right) = P\left( A|E \right)\]
\[ = \frac{P\left( A \right) \times P\left( E|A \right)}{P\left( A \right) \times P\left( E|A \right) + P\left( B \right) \times P\left( E|B \right) + P\left( C \right) \times P\left( E|C \right)}\]
\[ = \frac{\left( \frac{1}{3} \times 1 \right)}{\left( \frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{3}{5} \right)}\]
\[ = \frac{\left( \frac{1}{3} \right)}{\left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right)}\]
\[ = \frac{\left( \frac{1}{3} \right)}{\left( \frac{20 + 15 + 12}{60} \right)}\]
\[ = \frac{\left( \frac{1}{3} \right)}{\left( \frac{47}{60} \right)}\]
\[ = \frac{60}{3 \times 47}\]
\[ = \frac{20}{47}\]

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.