Question

Three fair coins numbered 1 and 0 are tossed simultaneously. Then variance Var (X) of the probability distribution of random variable X, where X is the sum of numbers on the uppermost faces, is

• A. 0.7
• B. 0.75
• C. 0.65
• D. 0.62

Hint:

Logic Tip: Tossing a coin with '0' and '1' faces is exactly a Bernoulli trial. For $n=3$ tosses with success probability $p=0.5$ (getting a '1'), the distribution is Binomial $B(3, 0.5)$. The variance of a binomial distribution is simply $npq = 3 \times 0.5 \times 0.5 = 0.75$. This shortcut bypasses all tabular calculations!

Answer 1

The Correct Option is B

Concept:
The random variable $X$ represents the sum of the numbers on the uppermost faces when three coins (each with faces 0 and 1) are tossed. This is equivalent to finding the number of '1's that appear. The variance of a probability distribution is calculated using the formula $Var(X) = E(X^2) - [E(X)]^2$, where $E(X) = \sum x_i P(x_i)$.
Step 1: Determine the sample space and the probability distribution of X.
When three fair coins are tossed, the sample space $S$ has $2^3 = 8$ outcomes: $$S = \{000, 001, 010, 100, 011, 101, 110, 111\}$$ The sum $X$ can take the values 0, 1, 2, or 3. The probabilities are: $$P(X=0) = \frac{1}{8} \quad (\text{for } 000)$$ $$P(X=1) = \frac{3}{8} \quad (\text{for } 001, 010, 100)$$ $$P(X=2) = \frac{3}{8} \quad (\text{for } 011, 101, 110)$$ $$P(X=3) = \frac{1}{8} \quad (\text{for } 111)$$
Step 2: Calculate the expected value E(X).
$$E(X) = \sum_{x=0}^{3} x \cdot P(X=x)$$ $$E(X) = 0\left(\frac{1}{8}\right) + 1\left(\frac{3}{8}\right) + 2\left(\frac{3}{8}\right) + 3\left(\frac{1}{8}\right)$$ $$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2} = 1.5$$
Step 3: Calculate E(X²) and the variance.
First, find $E(X^2)$: $$E(X^2) = \sum_{x=0}^{3} x^2 \cdot P(X=x)$$ $$E(X^2) = 0^2\left(\frac{1}{8}\right) + 1^2\left(\frac{3}{8}\right) + 2^2\left(\frac{3}{8}\right) + 3^2\left(\frac{1}{8}\right)$$ $$E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$$ Now, apply the variance formula: $$Var(X) = E(X^2) - [E(X)]^2$$ $$Var(X) = 3 - \left(\frac{3}{2}\right)^2 = 3 - \frac{9}{4}$$ $$Var(X) = \frac{12 - 9}{4} = \frac{3}{4} = 0.75$$