A player tosses 2 fair coins. He wins Rs. 5 if 2 heads appear, Rs. 2 if 1 head appear and Rs. 1 if no head appears, then variance of his winning amount is

Options

A. 6
B. `(5)/(2)`
C. `(9)/(4)`
D. `(17)/(2)`

Answer

C

Explanation

When player tosses 2 fair coins then
`S={HT,TH,T T,H H}`
Let X be a random variable that denotes the amount received by the player,
Then, X can take value 5,2 and 1. ,brgt Now, `P(X=5)=(1)/(4),P(X=2)=(2)/(4)=(1)/(2)`
and `P(X=1)=(1)/(4)`
Thus, the probability distribution of X is

`therefore`Variance of `X=sumX^(2)P(X)-[sumXP(X)]^(2)`
Now, `sumXP(X)=5xx(1)/(4)+2xx(1)/(2)+1xx(1)/(4)`
`=(5)/(4)+(4)/(4)+(1)/(4)=(10)/(4)=(5)/(2)`
`therefore`Variance off `X=sumX^(2)P(X)-(sumXP(X))^(2)`
`=(25xx(1)/(4)+4xx(1)/(2)+1xx(1)/(4))-((5)/(2))^(2)`
`=((25+8+1)/(4))-(25)/(4)=(9)/(4)`