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  • In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown, The man decided to throw a die thrice but to quit as and he gets a number greater than 4. Find the expected value of the amount he Wins/

In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown, The man decided to throw a die thrice but to quit as and he gets a number greater than 4. Find the expected value of the amount he Wins/loses.

 

 

 

 
 
 
 
 

Answers (1)

Let 'n' denote the number of throws required to get a number greater than 4 and X denotes the amount won/lost The man may get a number greater than 4 in the very first throw of the die or in second throw on is third throw.
Let p = probability of getting a number greater than 4
 = \frac{2}{6}
q=1-p= \frac{4}{6} thus we have the following distribution for x.
Number of thrown (n)  1         2          3      3
Amount lost(x)             5         4          3      -3
Probability(p(x))          \left [ \frac{2}{6} \right ]   \left [ \frac{4}{6}\times \frac{2}{6} \right ]\left [ \frac{4}{6}\times \frac{4}{6}\times \frac{2}{6} \right ]\left [ \frac{4\times4\times4}{6\times6\times6} \right ]
E\left ( x \right )= 5\times\frac{2}{6}+4\times\frac{4}{6}\times\frac{2}{6}+3\times\frac{4}{6}\times\frac{4}{6}\times\frac{2}{6}+\left ( -3 \right )\times\frac{4}{6}\times\frac{4}{6}\times\frac{4}{6}
            = \frac{Rs\, 19}{9} is the expected amount he could win 

Posted by

Ravindra Pindel

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