Question

A coin is tossed twice. Let $X$ be a random variable defined as the number of heads minus the number of tails. Obtain the probability distribution of $X$ and also find its mean.

Hint:

For random variables with discrete outcomes, the expected value (mean) is calculated by summing the products of each outcome and its probability.

Answer 1

Possible outcomes when tossing a coin twice: \[ \{HH, HT, TH, TT\}. \] Let $X$ be the number of heads minus the number of tails. We compute $X$ for each outcome: - For $HH$, $X = 2 - 0 = 2$.
- For $HT$, $X = 1 - 1 = 0$.
- For $TH$, $X = 1 - 1 = 0$.
- For $TT$, $X = 0 - 2 = -2$.
The probability distribution is:
- $P(X = 2) = \frac{1}{4}$ (since only one outcome has 2 heads).
- $P(X = 0) = \frac{2}{4} = \frac{1}{2}$ (since two outcomes have 1 head and 1 tail).
- $P(X = -2) = \frac{1}{4}$ (since only one outcome has 2 tails).
Thus, the probability distribution is: \[ P(X = 2) = \frac{1}{4}, \quad P(X = 0) = \frac{1}{2}, \quad P(X = -2) = \frac{1}{4}. \] To find the mean (expected value) of $X$, we use: \[ E(X) = \sum x \cdot P(X = x) = 2 \times \frac{1}{4} + 0 \times \frac{1}{2} + (-2) \times \frac{1}{4} = \frac{2}{4} + 0 - \frac{2}{4} = 0. \] Thus, the mean of $X$ is $E(X) = 0$.