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  • 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

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}
 

Posted by

Ravindra Pindel

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