# Find the Mean and Variance of the Number of Tails in Three Tosses of a Coin. - Mathematics

Sum

Find the mean and variance of the number of tails in three tosses of a coin.

#### Solution

Let X denote the number of tails in three tosses of a coin.
Then, X can take the values 0, 1, 2 and 3.
Now,

$P\left( X = 0 \right) = P\left( HHH \right) = \frac{1}{8},$

$P\left( X = 1 \right) = P\left( \text{ THH or HHT or HTH }\right) = \frac{3}{8},$

$P\left( X = 2 \right) = P\left(\text{ TTH or THT or HTT }\right) = \frac{3}{8},$

$P\left( X = 3 \right) = P\left( TTT \right) = \frac{1}{8}$

Thus, the probability distribution of X is given by

 x P(X) 0 $\frac{1}{8}$ 1 $\frac{3}{8}$ 2 $\frac{3}{8}$ 3 $\frac{1}{8}$

Computation of mean and variance

 xi pi pixi pixi2 0 $\frac{1}{8}$ 0 0 1 $\frac{3}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ 2 $\frac{3}{8}$ $\frac{6}{8}$ $\frac{12}{8}$ 3 $\frac{1}{8}$ $\frac{3}{8}$ $\frac{9}{8}$ ∑pixi = $\frac{3}{2}$ ∑pixi2=3

$\text{ Mean } = \sum p_i x_i = \frac{3}{2}$
$\text{ Variance } = \sum p_i {x_i}^2 - \left( \text{ Mean} \right)^2$
$= 3 - \left( \frac{3}{2} \right)^2$
$= 3 - \frac{9}{4}$
$= \frac{3}{4}$

Is there an error in this question or solution?
Chapter 32: Mean and Variance of a Random Variable - Exercise 32.2 [Page 43]

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 32 Mean and Variance of a Random Variable
Exercise 32.2 | Q 4 | Page 43
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