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: Probability of each outcome:
Two fair coins: P(2 heads) \(=\dfrac14\), P(1 head) \(=\dfrac12\), P(no head) \(=\dfrac14\).

Step 2: Winning amount X:
\(X=5\) with probability \(\dfrac14\), \(X=3\) with probability \(\dfrac12\), \(X=2\) with probability \(\dfrac14\).

Step 3: Mean and mean of squares:
\(E(X)=\dfrac54+\dfrac32+\dfrac24=3.25\). \(E(X^2)=\dfrac{25}{4}+\dfrac92+\dfrac44=11.75\).

Step 4: Variance:
\(\text{Var}(X)=11.75-(3.25)^2=11.75-10.5625=1.1875\). Option A.

Step 5: Why the other options are wrong.
1.8175, 1.1785 and 1.8157 are digit-shuffled values that come from small arithmetic slips.

Final Answer:
The variance is 1.1875. \[ \boxed{\text{(A) }1.1875} \]