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

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 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?

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Answer

Let $E_{1 }= a$ two headed coin, $E_{2 }= a$ biased coin, $E_{3 }= an$ unbiased coin and $A = A $ head is shown
$\text{Now}\ \ \text{P}(\text{E}_1)=\frac{1}{3},\ \text{P}(\text{E}_2)=\frac{1}{3},\ \text{P}(\text{E}_3)=\frac{1}{3}$
$\text{P}(\text{A}|\text{E}_1)=1,\ \text{P}(\text{A}|\text{E}_2)=\frac{75}{100}=\frac{3}{4}\ \text{and}\ \text{P}(\text{A}|{\text{E}_3})=\frac{1}{2}$
Therefore, by Bayes’ theorem,
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{A}|\text{E}_3)}$
$=\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}$
$=\frac{{4}}{{1}+\frac{3}{4}+\frac{1}{2}}$
$=\frac{4}{4+3+2}$
$=\frac{4}{9}$

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