Question

A player tosses two fair coins, he wins Rs. \(5\) if two heads appear, Rs. \(3\) if one head appears and Rs. \(2\) if no head appears, then the variance of the winning amount is

• A. \(1.1875\)
• B. \(1.8175\)
• C. \(1.1785\)
• D. \(1.8157\)

Hint:

Find the distribution of the winnings, then use Var = E(X^2) - (E X)^2.

Answer 1

The Correct Option is A

Step 1: Shift the values to simplify:
Subtract 3 from each amount; variance is unchanged. Values: $+2,\ 0,\ -1$ with probabilities $\dfrac14,\dfrac12,\dfrac14$.

Step 2: Compute with shifted values:
Mean $=\dfrac24+0-\dfrac14=0.25$. Mean of squares $=\dfrac44+0+\dfrac14=1.25$.

Step 3: Variance:
$1.25-0.25^2=1.25-0.0625=1.1875$. Option A.

Final Answer:
Using shifted values 2, 0, -1 gives variance 1.25 - 0.0625 = 1.1875. \[ \boxed{\text{(A) }1.1875} \]