Question

A player tosses 2 fair coins. He wins ₹ 5 if 2 heads appear, ₹ 21f 1 head appears a) if no heads appear, then the variance of his winning amount is

• A. 9/4

Answer 1

Answer: (A)

9/4

Answer 2

Let the winning amounts be defined by:

$$X = \begin{cases} 5 & \text{if 2 heads (prob. }1/4\text{)}\\[1mm] 2 & \text{if 1 head (prob. }1/2\text{)}\\[1mm] 1 & \text{if 0 heads (prob. }1/4\text{)} \end{cases}$$

1. Expected Value:

$$E[X] = \frac{1}{4}(5) + \frac{1}{2}(2) + \frac{1}{4}(1)= \frac{5}{4} + 1 + \frac{1}{4} = \frac{5+4+1}{4} = \frac{10}{4} = 2.5.$$

2. Second Moment:

$$E[X^2] = \frac{1}{4}(25) + \frac{1}{2}(4) + \frac{1}{4}(1)= \frac{25}{4} + 2 + \frac{1}{4} = \frac{25+8+1}{4} = \frac{34}{4}=8.5.$$

3. Variance:

$$Var(X)=E[X^2]-(E[X])^2= 8.5 - (2.5)^2 = 8.5 - 6.25 = 2.25 = \frac{9}{4}.$$

Thus, the variance of his winning amount is $\frac{9}{4}$.