Random Variables and Probability Distributions

LEARNING OBJECTIVES

1. To learn the concept of the probability distribution of a discrete random variable.
2. To learn the concepts of the mean, variance, and standard deviation of a discrete random variable, and how to compute them.

Probability Distributions

Associated to each possible value $x$ of a discrete random variable $X$ is the probability $P(x)$ that $X$ will take the value $x$ in one trial of the experiment.

Definition

The probability distribution of a discrete random variable $X$ is a list of each possible value of $X$ together with the probability that $X$ takes that value in one trial of the experiment.

The probabilities in the probability distribution of a random variable $X$ must satisfy the following two conditions:

1. Each probability $P(x)$ must be between o and 1: $0 \leq P(x) \leq 1$.

2. The sum of all the probabilities is $1: \Sigma P(x)=1$.

EXAMPLE 1

A fair coin is tossed twice. Let $X$ be the number of heads that are observed.

a. Construct the probability distribution of $X$.

b. Find the probability that at least one head is observed.

Solution:

$\begin{align*} P(X \geq 1)=P(1)+P(2)=0.50+0.25=0.75 \end{align*}$

Figure 4.1

Probability Distribution for Tossing a Fair Coin Twice

EXAMPLE 2

A pair of fair dice is rolled. Let $X$ denote the sum of the number of dots on the top faces.

a. Construct the probability distribution of $X$.

b. Find $P(X \geq 9)$.

c. Find the probability that $X$ takes an even value.

Solution:

The sample space of equally likely outcomes is

a. The possible values for $X$ are the numbers 2 through 12 . $X=2$ is the event $\{11\}$, so $P(2)=1 / 36 . X=3$ is the event $\{12,21\}$, so $P(3)=2 / 36$. Continuing this way we obtain the table

$\begin{array}{c|ccccccccccc} x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(x) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \end{array}$

This table is the probability distribution of $X$.

b. The event $X \geq 9$ is the union of the mutually exclusive events $X=9, X=10, X=11$, and $X=12$. Thus

$\begin{align*} P(X \geq 9)=P(9)+P(10)+P(11)+P(12)=\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{10}{36}=0.2 \overline{7} \end{align*}$

c. Before we immediately jump to the conclusion that the probability that $x$ takes an even value must be $0.5$, note that $X$ takes six different even values but only five different odd values. We compute

 $\begin{align*} \begin{aligned} P(X \text { is even }) &=P(2)+P(4)+P(6)+P(8)+P(10)+P(12) \\ &=\frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36}=\frac{18}{36}=0.5 \end{aligned} \end{align*}$

A histogram that graphically illustrates the probability distribution is given in Figure 4.2

"Probability Distribution for Tossing Two Fair Dice".

Figure 4.2

Probability Distribution for Tossing Two Fair Dice