A player tosses two coins. He wins ₹10 if 2 heads appear, ₹5 if one head appears, and ₹2 if no head appears. Then variance of winning amount is ______.
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Always organize random variable problems into a clean table with columns for $x$, $P(x)$, $x \cdot P(x)$, and $x^2 \cdot P(x)$. Summing the last two columns instantly gives you the required components for the variance formula.
Step 1: Understanding the Question:
We must construct a probability distribution for a random variable $X$ representing the "winning amount" from tossing two coins, and then calculate the statistical variance of $X$. Step 2: Key Formula or Approach:
The variance of a discrete random variable is given by:
$$\text{Var}(X) = E(X^2) - [E(X)]^2$$
Where $E(X) = \sum x_i P(x_i)$ is the Expected Value (mean), and $E(X^2) = \sum x_i^2 P(x_i)$. Step 3: Detailed Explanation:
1. Establish the sample space and probabilities for tossing 2 coins:
Sample Space = $\{HH, HT, TH, TT\}$ (Total 4 outcomes).
- Probability of 2 Heads ($HH$): $1/4$
- Probability of 1 Head ($HT, TH$): $2/4 = 1/2$
- Probability of 0 Heads ($TT$): $1/4$
2. Map these to the winning amounts ($X$):
- $X = 10$ with $P(X) = 1/4$
- $X = 5$ with $P(X) = 1/2$
- $X = 2$ with $P(X) = 1/4$
3. Calculate the Expected Value $E(X)$:
$$E(X) = 10(1/4) + 5(1/2) + 2(1/4)$$
$$E(X) = 2.5 + 2.5 + 0.5 = 5.5$$
4. Calculate $E(X^2)$:
$$E(X^2) = (10^2)(1/4) + (5^2)(1/2) + (2^2)(1/4)$$
$$E(X^2) = 100(0.25) + 25(0.5) + 4(0.25)$$
$$E(X^2) = 25 + 12.5 + 1 = 38.5$$
5. Calculate Variance $\text{Var}(X)$:
$$\text{Var}(X) = 38.5 - (5.5)^2$$
$$\text{Var}(X) = 38.5 - 30.25 = 8.25$$ Step 4: Final Answer:
The variance is 8.25, matching option (b).
