Q. 11  In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins / loses.

Answers (1)
S seema garhwal

In a throw of die,

probability of getting six  = P

                   P=\frac{1}{6}

probability of not getting six  = q

             q=1-P=1-\frac{1}{6}=\frac{5}{6}

There are three cases :

1. Gets six in the first throw, required probability is \frac{1}{6}

     The amount he will receive is Re. 1

2.. Does not gets six in the first throw and gets six in the second throw, then the probability 

                                                                         =\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}

               

     The amount he will receive is - Re.1+ Re.1=0

3. Does not gets six in first 2 throws and gets six in the third throw, then the probability 

                                                                      =\frac{5}{6}\times\frac{5}{6}\times \frac{1}{6}=\frac{25}{216}            

     Amount he will receive is -Re.1 - Re.1+ Re.1= -1

Expected value he can win  :

                                               =1\times \frac{1}{6}+0\times \frac{5}{36}+(-1)\times \frac{25}{216}

                                             = \frac{1}{6}-\frac{25}{216}

                                             = \frac{11}{216}

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