Probability Distributions for Discrete Random Variables (College Board AP® Statistics)

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Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Discrete probability distributions

What is a random variable?

A random variable is a variable whose values represent the outcomes of a random event
- e.g. a random variable could represent the height of a person or the number of hats owned by a person
The word random does not refer to the probability of each outcome occurring
- It means the variable can take any value from its possible outcomes
- The outcome is not pre-determined
Random variables are denoted by uppercase letters ( $X$ , $Y$ , etc)
The outcomes are denoted by lowercase letters ( $x$ , $y$ , etc)
The probability that the random variable $X$ takes the value $x$ is denoted as $P (X = x)$ or $P (x)$

What is a discrete random variable?

A discrete random variable is a random variable with a countable number of values
- The number of values could be finite
  - e.g. the number of times a coin lands on heads when flipped 10 times,
    the outcomes are {1, 2, 3, ..., 10}
- The number of values could be infinite
  - e.g. the number of times a coin is flipped until it first lands on heads,
    the outcomes are {1, 2, 3, ...}

What is a discrete probability distribution?

A discrete probability distribution describes the probabilities of the occurrences of each possible outcome associated with a discrete random variable
A discrete probability distribution can be given in several forms:
- a table
- a function
- or a graph

A diagram showing a scenario of an eight-sided spinner with numbers 0, 1/3, 5, -2, and the probability distribution of the scenario in three forms. A table showing probabilities of the spinner landing on each number, P(0)=1/8, P(1/3)=1/8, P(5)=1/2, P(-2)=1/4; a function describing the same probabilities and a bar graph of these values. — **Example of the ways to display a probability distribution for a discrete random variable**

You can interpret a probability distribution in its graphical form by looking at:
- its shape
- its center
- and its spread
If the outcomes are equally likely then the bars on the graph have the same heights
- These are called discrete uniform distributions

How can I use a discrete probability distribution?

The sum of all the probabilities is equal to 1
- You can use this to find a missing probability
To find the probability of an event
- Identify the outcomes which are in the event
- Add together the probabilities of those outcomes

Examiner Tips and Tricks

If you introduce a random variable when answering an exam question, make sure you state what the variable represents.

For example, let the random variable $X$ represent the number of points scored in an exam.

Worked Example

Each day, Josh drives through four sets of traffic lights on his way to work. The random variable, $X$ , represents the number of sets of traffic lights that Josh has to stop at on his way to work on a randomly selected day. The table below shows the probability distribution of $X$ .

$x$	0	1	2	3	4
$P (X = x)$	0.4	0.3	$P (X = 2)$	0.1	0.05

(a) Calculate the probability that Josh has to stop for exactly 2 sets of traffic lights on his way to work on a randomly selected day.

Answer:

The required probability is $P (X = 2)$

The probabilities of the outcomes add up to 1

Form an equation and solve it

$\begin{array}{rcl} 0.4 + 0.3 + P (X = 2) + 0.1 + 0.05 & = & 1 \\ P (X = 2) + 0.85 & = & 1 \\ P (X = 2) & = & 0.15 \end{array}$

The probability that Josh has to stop for exactly 2 sets of traffic lights on his way to work on a randomly selected day is 0.15

(b) Calculate the probability that Josh has to stop for at least one set of traffic lights on his way to work on a randomly selected day.

Answer:

The required probability is $P (X \geq 1)$

At least once means Josh could stop at 1, 2, 3 or 4 traffic lights

This is all of the outcomes except 0

Either add the probabilities for $X = 1, 2, 3, 4$ or subtract the probability for $X = 0$ from 1

$\begin{array}{rcl} P (X \geq 1) & = & P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) \\ = & 0.3 + 0.15 + 0.1 + 0.05 \\ = & 0.6 \end{array}$

$\begin{array}{rcl} P (X \geq 1) & = & 1 - P (X = 0) \\ = & 1 - 0.4 \\ = & 0.6 \end{array}$

The probability that Josh has to stop for at least one set of traffic lights on his way to work on a randomly selected day is 0.6

Last updated: 13 August 2024

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