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# 4.2: Probability Distribution for a Discrete Random Variable

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Know and understand the notion of discrete random variables.
• Learn how to use discrete random variables to solve probabilities of outcomes.

Now, we want to specify the possible values that a discrete random variable can assume. The example below illustrates how.

Example:

Suppose you simultaneously toss two fair coins. Let x\begin{align*}x\end{align*} be the number of heads observed. Find the probability associated with each value of the random variable x\begin{align*}x\end{align*}.

Solution:

Since there are two coins and each coin can be either heads or tails, there are four possible outcomes (HH,HT,TH,TT)\begin{align*}(HH, HT, TH, TT)\end{align*} each with probability 1/4\begin{align*}1/4\end{align*}. Since x\begin{align*}x\end{align*} is the number of heads observed, then x=0,1,2\begin{align*}x = 0, 1, 2\end{align*}. The Venn diagram below shows the two-coin experiment.

From the Venn diagram, we can identify the probabilities of the simple events associated with each value of x\begin{align*}x\end{align*}:

P(x=0)P(x=1)P(x=2)=P(TT)=1/4=P(HT)+P(TH)=1/4+1/4=1/2=P(HH)=1/4\begin{align*}P(x = 0) & = P(TT) = 1/4 \\ P(x = 1) & = P(HT) + P(TH) = 1/4 + 1/4 = 1/2 \\ P(x = 2) & = P(HH) = 1/4\end{align*}

Thus, we have just had a complete description of the values of the random variables and have calculated the associated probabilities that are distributed over these values. We refer to it as the probability distribution. This probability distribution can be represented in different ways, sometimes in a tabular form and sometimes in a graphical one. Both forms are shown below.

In tabular form,

x012P(x)1/41/21/4\begin{align*}x && P(x)\\ 0 && 1/4 \\ 1 && 1/2 \\ 2 && 1/4\end{align*}

Figure: The Tabular Form of the Probability Distribution for the Random Variable in the First Example.

In a graphical form,

We can also describe the probability distribution in a mathematical formula, which we will do later in the chapter.

## Probability Distribution

The probability distribution of a discrete random variable is a graph, a table, or a formula that specifies the probability associated with each possible value that the random variable can assume.

Two conditions must be satisfied for all probability distributions:

Two conditions must be satisfied for the probability distribution:

P(x)0\begin{align*}P(x) \ge 0\end{align*}, for all values of x\begin{align*}x\end{align*}

xP(x)=1\begin{align*}\sum_x P(x) = 1\end{align*}

The symbol xP(X)\begin{align*}\sum_x P(X)\end{align*} means "add the values of P(x)\begin{align*}P(x)\end{align*} for all values of x\begin{align*}x\end{align*}"

Example:

What is the probability distribution of the number of yes votes for three voters (see the first example in the first section Introduction)

Solution:

Since each of the 8\begin{align*}8\end{align*} outcomes is equally likely, the following table gives the probability of each value of the random variable.

Value of Random Variable

(number of Yes votes) Probability

Probability
3\begin{align*}3\end{align*} 1/8=0.125\begin{align*}1/8 = 0.125\end{align*}
2\begin{align*}2\end{align*} 3/8=0.375\begin{align*}3/8 = 0.375\end{align*}
1\begin{align*}1\end{align*} 3/8=0.375\begin{align*}3/8 = 0.375\end{align*}
1/8=0.125\begin{align*}1/8 = 0.125\end{align*}

Figure: Tabular Representation of the Probability Distribution for the Random Variable in this Example.

The table and the graph are two ways to show the probability distribution. Note that the graph of a probability distribution can be either a line graph or a bar graph.

## Lesson Summary

1. The probability distribution of a discrete random variable is a graph, a table, or a formula that specifies the probability associated with each possible value that the random variable can assume.

2. All probability distributions must satisfy:

P(x)>0\begin{align*}P(x) > 0\end{align*} (for all values of x\begin{align*}x\end{align*})

xP(x)=1\begin{align*}\sum_x P(x) = 1\end{align*}

## Review Questions

1. Consider the following probability distribution: xp(x)40.100.310.430.2\begin{align*}& x && -4 && 0 && 1 && 3 \\ & p(x) && 0.1 && 0.3 && 0.4 && 0.2\end{align*}
1. What are all the possible values of x?\begin{align*}x?\end{align*}
2. What value of x\begin{align*}x\end{align*} is most likely to happen?
3. What is the probability that x\begin{align*}x\end{align*} is greater than zero?
4. What is the probability that x=2?\begin{align*}x = -2?\end{align*}
2. A fair die is tossed twice and the up face is recorded. Let x\begin{align*}x\end{align*} be the sum of the up faces.
1. Give the probability distribution of x\begin{align*}x\end{align*} in tabular form.
2. What is p(x8)?\begin{align*}p(x \ge 8)?\end{align*}
3. What is p(x<8)?\begin{align*}p(x < 8)?\end{align*}
4. What is the probability that x\begin{align*}x\end{align*} is odd? Even?
5. What is p(x=7)?\begin{align*}p(x = 7)?\end{align*}
3. If a couple has three children, what is the probability that they have at least one boy?

1. x={4,0,1,3};\begin{align*}x = \left \{-4, 0, 1, 3\right \};\end{align*}
2. 1;\begin{align*}1;\end{align*}
3. 0.6;\begin{align*}0.6;\end{align*}
1. xp(x)21/3632/3643/3654/3665/3676/3685/3694/36103/36112/36121/36\begin{align*}& x && 2 && 3 && 4 && 5 && 6 && 7 && 8 && 9 && 10 && 11 && 12 \\ & p(x) && {1/36} && {2/36} && 3/36 && 4/36 && 5/36 && 6/36 && 5/36 && 4/36 && 3/36 && 2/36 && 1/36\end{align*}
2. 15/36;\begin{align*}15/36;\end{align*}
3. 21/36;\begin{align*}21/36;\end{align*}
4. 18/36,18/36;\begin{align*}18/36, 18/36;\end{align*}
5. 6/36\begin{align*}6/36\end{align*}
1. 7/8\begin{align*}7/8\end{align*}

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