# 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.

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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