If the probability distribution table for no. of “Heads” in three throws of coin is
| X | 0 | 1 | 2 | 3 |
| P(X = x) | \(\displaystyle\frac{1}{8}\) | \(\displaystyle\frac{3}{8}\) | \(\displaystyle\frac{3}{8}\) | \(\displaystyle\frac{1}{8}\) |
Then Var (X) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{4}\)
The variance is \(\mathrm{Var}(X) = E(X^2) - [E(X)]^2\).
\( \displaystyle E(X) = 0\times\frac{1}{8} + 1\times\frac{3}{8} + 2\times\frac{3}{8} + 3\times\frac{1}{8} = \frac{12}{8} = \frac{3}{2} \)
\( \displaystyle E(X^2) = 0\times\frac{1}{8} + 1\times\frac{3}{8} + 4\times\frac{3}{8} + 9\times\frac{1}{8} = \frac{24}{8} = 3 \)
\( \displaystyle \mathrm{Var}(X) = 3 - \frac{9}{4} = \frac{3}{4} \)
Shortcut: The number of heads in 3 tosses of a fair coin is binomial with n = 3, p = q = 1/2, so
\( \displaystyle \mathrm{Var}(X) = npq = 3\times\frac{1}{2}\times\frac{1}{2} = \frac{3}{4} \)
NCERT Exemplar:
Probability
