There are three coins. One is a coin having tails on both faces, another is a biased coin that comes up tails  70% of the time and the third is an unbiased coin. One of the coins is chosen at random and tossed, it shows tail. Find the probability that it was a coin with tail on both the faces.

 

 

 

 
 
 
 
 

Answers (1)

Let E: the coin shows tail
Also let E_{1},E_{2},E_{3}, : the related coin is having tails on both the faces, a baised coin & un baised coin respectively.
Here p\left ( E_{1} \right )= p\left ( E_{2} \right )= p\left ( E_{3} \right )= \frac{1}{3}
         p\left ( \frac{E}{E_{1}} \right )= 1,p\left ( \frac{E}{E_{2}} \right )= 70/100p\left ( \frac{E}{E_{1}} \right )= 1,p\left ( \frac{E}{E_{2}} \right )= 70\: ^0/_0, p\left ( \frac{E}{E_{3}} \right )= \frac{1}{2}
By Baye's Theorem p\left ( \frac{E_{1}}{E} \right )= \frac{p\left ( \frac{E}{E_{1}} \right )p\left ( E_{1} \right )}{p\left ( \frac{E}{E_{1}} \right )p\left ( E_{1} \right )+p\left ( \frac{E}{E_{2}} \right )p\left ( E_{2} \right )+p\left ( \frac{E}{E_{3}} \right )p\left ( E_{3} \right )}
\Rightarrow p\left ( \frac{E_{1}}{E} \right )= \frac{1\times \frac{1}{3}}{1\times \frac{1}{3}+\frac{70}{100}\times \frac{1}{3}+\frac{1}{2}\times \frac{1}{3}}
                        \Rightarrow \frac{10}{22}\; or\; \frac{5}{11}
 

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