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In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs.50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is : 

 

  • Option 1)

    0

     

     

     

  • Option 2)

    \frac{400}{3}gain

  • Option 3)

    \frac{400}{9}loss

  • Option 4)

    \frac{400}{3}loss

Answers (1)

best_answer

 

Multiplication Theorem of Probability -

If A and B are any two events then 

P\left ( A\cap B \right )= P\left ( B \right )\cdot P\left ( \frac{A}{B} \right )


 

- wherein

where B\neq \phi

 

 

The law of Total Probability -

Let S be the sample space and E1, E2, ......En be n mutually exclusive and exhaustive events associated with a random experiment.

- wherein

\Rightarrow P\left ( A \right )= P\left ( A\cap E_{1} \right )+P\left ( A\cap E_{2} \right )+\cdots +P\left ( A\cap E_{n} \right )

\Rightarrow P\left ( A \right )= P\left ( E_{1} \right )\cdot P\left ( \frac{A}{E_{1}} \right )+P\left ( E_{2} \right )\cdot P\left ( \frac{A}{E_{2}} \right )+\cdots P\left ( E_{n} \right )\cdot P\left ( \frac{A}{E_{n}} \right )

where A is any event which occurs with E1, E2, E3......En.

 

p (success)=   p ( 5 or 6 )  =1/3

expectation \: =100\times \frac{1}{3}+100\times \frac{1}{9}-\frac{400}{9}\\\\\\= 0


Option 1)

0

 

 

 

Option 2)

\frac{400}{3}gain

Option 3)

\frac{400}{9}loss

Option 4)

\frac{400}{3}loss

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