Have you ever played an archery game and tried to see how many times you can throw an arrow before hitting a particular target? When you do this, you are testing the probabilities and outcomes of random events. We can express and describe the outcomes of random events with random variables.
Discrete random variables are a type of random variable in which values are finite.
In other words, values are countable and have a limited number of outcomes. Examples of discrete random variables are the number of books in a pack, the number of cubes of sugar in a box, the number of goats in a pen, and a person’s shoe size, among others. In this lesson, we are going to learn in detail about discrete random variables and their probability distributions.
A discrete random variable is a variable that may take on only a limited number of specified, countable values. All values within the random variable's domain have probabilities associated with them. These probabilities must sum up to \(1\) when all possible values are considered.
Discrete random variables: Distribution types
First, let's recall the concept of distribution. The probability distribution for a discrete random variable X is a comprehensive set of each potential value of \(X\), along with the likelihood that \(X\) will take that value in one trial of the experiment. In other words, discrete probability distributions are used to describe the probabilities associated with the discrete random variable's values. Two common types of discrete random variables are binomial random variables (with a binomial probability distribution) and geometric random variables (with a geometric probability distribution).
In this article, we consider only binomial and geometric random variables, which are relevant for an AP Statistics course. Other types which will not be covered in this article include Bernoulli, Multinomial, Hypergeometric, and Poisson distributions.
Binomial and geometric probability distributions of discrete random variables
The probability distribution of a discrete random variable refers to the catalog of the potential values of that discrete random variable, along with the probability that it will take that value in one try of the experiment.
The distributions of discrete random variables must satisfy the following two conditions given a discrete random variable \(X\):
Each probability \(P(x)\) must be between \(0\) and \(1, 0 \leq P(x) \leq 1\).
The addition of all probabilities does not exceed \(1\): \(\sum P(x) = 1\).
Let's take a look at an example of what is meant by a probability distribution of a discrete random variable.
Let \(X\) be the number of heads to be observed from tossing a fair coin twice. First, construct the probability distribution of \(X\). Second, find the probability that at least one heads is observed.
Solution:
For this sample space, the possible values of \(X\) are \(0\), \(1\), and \(2\). The potential outcomes have equal chances of occurring and follow as:
\(S = {hh, ht, th, tt}\).
That is, "\(hh\)" refers to the outcome of two heads,
"\(ht\)" refers to the outcome of one head and one tail, and so on.
As the number of heads observed is represented by \(X = 0\):
\(X = 0\) corresponds to \({tt}\), with no heads observed
\(X = 1\) corresponds to \({ht, th}\), with \(1\) heads observed
\(X = 2\) corresponds to \({hh}\), with \(2\) heads observed
By simply counting, we derive the probability of each of these three events, as represented by the discrete variable \(X\). Thus:
Table 1: Probability Distribution of Tossing a Fair Coin Twice
| \(X\) | \(0\) | \(1\) | \(2\) |
| \(P(x)\) | \(0.25\) | \(0.50\) | \(0.25\) |
The probability that at least one head is observed is an event that can be described by the mathematical expression: \(X \geq 1\). The probability of this particular event (at least one head) is calculated by the addition of the two mutually exclusive events of \(X =1\) and \(X = 2\). Therefore, \(P (X \geq 1) = P (1) + P (2) = 0.50 + 0.25 = 0.75\). In other words, there is a \(75\%\) chance that at least one heads will result from tossing a coin twice.
Binomial random variables
A binomial random variable is a type of discrete random variable which we use to express the frequency of a particular outcome (or event) throughout a fixed number of experimental trials. The binomial random variable is expressed within a binomial distribution.
In order for a discrete random variable to also be a binomial random variable, the following characteristics must apply:
The number of trials is predetermined or fixed.
The trials are independent. (The trials' results do not impact one another.)
For each trial, only two outcomes may occur: a “success” or a “failure.” In other words, the particular event of interest will either happen or it will not happen. This type of result can also be called "binary."
Any given trial has the same probability of "success" as the others in the experiment.
If the discrete random variable \((X)\) is classified as binomial, it can be used to count the number of successes in the n trials. This implies that \(X\) has a binomial distribution with the following two parameters:
"\(n\)," which measures the number of trials and
"\(p\)," which measures the probability of success of a particular event.
For example, let's consider a random sample of 125 nurses selected from a large hospital in which the proportion of nurses who are female is 57%. Suppose X denotes the number of female nurses in the sample. In this experiment, there are 125 (n = 125) identical and independent trials of a common procedure: selecting a nurse at random. There are exactly two possible outcomes for each trial, “success” (the event that we are counting, that the nurse is female) and “failure” (not female). Finally, the probability of success on any one trial is the same number (p = 0.57). As the proportion of nurses is 57% female, a random selection would therefore provide a 57% chance of selecting a female nurse. Thus, X is a binomial random variable with parameters n = 125 and p = 0.57.
Probability Formula for a Binomial Random Variable
If a distribution is described by a binomial random variable, you may apply the formula below to calculate the probability of \(X\):
\[P(X=x)= \begin{pmatrix} n \\ X \end {pmatrix} p^x q^{n-x}={\frac{n!}{(n-X)!}\cdot (X!)p^x q^{n-x}}\]
where,
\(P\) = binomial probability
\(x\) = frequency of specific outcome within a specific number of trials
\(\begin{pmatrix} n \\ X \end {pmatrix} \)= number of combinations
\(p\) = probability of success on a single trial
\(q\) = probability of failure on a single trial
\(n\) = number of trials
Assuming a fair coin is tossed \(10\) times, what is the probability of getting \(6\) tails?
Solution
There are two outcomes that can be obtained in a coin toss experiment: a heads or a tails. Therefore:
The probability of getting a tails is \(50\%\) (or \(0.5\)) in a given toss.
Number of trials, \(n = 10\).
Frequency of outcome, \(x = 6\).
- Probability of success on a single trial, \(p = 0.5\)
- Probability of failure on a single trial, \(q = 0.5\).
Plugging these values into the formula:
\[P(X=x)={\frac{n!}{(n-X)!}\cdot (X!)p^x q^{n-x}}\]
\[P(X=x)=0.205\]
Geometric random variables
Geometric random variables are discrete random variables that form a geometric distribution. This concept is used in several spheres of life such as cost-benefit analysis in financial industries, among others.
The experimental conditions required for geometric random variables are very similar to those of binomial random variables: they both categorize trials as either successes or failures, and the trials must be independent, with the same probability of occurrence for each. However, unlike binomial random variables, the number of trials are not fixed for geometric random variables beforehand. Rather, it depends on the number of successive failures that occur before a success is achieved.
For example, consider a geometric random variable, \(X = 3\), which represents obtaining a number \(3\) as the result of the roll of a fair die.
In this geometric random variable experiment, we would count the number of times the die is rolled before a value of \(3 (X = 3)\) is achieved once.
The trial in which a \(3\) is rolled is labeled as a "success," and any trial in which a \(3\) is not rolled is labeled as a "failure." As this is a geometric random variable experiment, we only need to obtain one success in order to finish it.
Since the observations are independent of each other, the probability that \(X = 3\) (a \(3\) will result from the roll of the die) will be \(1/6\) for each roll. This probability of \(1/6\) is because the die has six sides, which gives values of 1 through \(6\).
Probability Formula for a Geometric Random Variable
If a distribution is described by a geometric random variable, you may apply the formula below to calculate the probability of \(X\):
\[P(X=x)=(1-p)^{x-1}p\]
where \(0<p<1\) and \(x=1, 2, 3...\)
A representative from the National Theatre Marketing Division randomly selects people on a random street in Washington D.C. until he finds a person who attended the last movie show.
Let \(p\), the probability that he succeeds in finding such a person, equal \(0.20\). And, let \(X\) denote the number of people he selects until he finds his first success.
a. What is the probability that the marketing representative must select 4 people before he finds one who attended the movie show?
Solution:
\[f(x)=P(X=x)=(1-p)^{x-1}p\]
\[P(X=4)=(1-0.2)^{4-1}0.2=0.1024\]
Therefore, there is approximately a \(10\%\) chance that the marketing representative would have to select \(4\) people before he finds one who attends the last movie show.
The mean, variance, and standard deviation of discrete random variables
In this section, we discuss the mean, variance, and standard deviation as applied to discrete random variables. Then, we apply these concepts to an example problem.
Mean
The mean is also known as the expected value, and it refers to the average of the values. For discrete random variables, the mean refers to the average of all values as assigned to events that occur in repeated trials of the experiment. The mean of a discrete random variable is given by the expression below:
\[\mu= E(x)=\sum x \cdot P(x)\]
Thus, the mean is derived by multiplying each value by its probability of occurring. These values are then summed up to generate the mean of the experiment.
Find the mean of the discrete probability distribution below:
\(x\) | \(-2\) | \(1\) | \(2\) | \(3.5\) |
\(P(x)\) | \(0.21\) | \(0.34\) | \(0.54\) | \(0.31\) |
Solution:
Following the formula:
\[\mu= E(x)=\sum x \cdot P(x)\]
\[\mu = ((-2)\cdot(0.21))+((1)\cdot(0.34))+((2)\cdot(0.54))+((3.5)\cdot(0.31))\]
Thus, \(\mu = 2.085\)
Variance
The variance measures how spread out the data is. More specifically, it is the weighted average measuring the squared deviations or variabilities of each value about the mean of repeated trials of an experiment. This is given by:
\[\sigma^2 = \sum (x-\mu)^2P(x)\]
Standard Deviation
Similar to the variance, the standard deviation also measures the data's dispersion. Specifically, it measures the magnitude by which each observation deviates from the mean. To compute the standard deviation of a discrete random variable, simply take the square root of the value of the variance.
\[\sigma=\sqrt{\sum(x-\mu)^2P(x)}\]
A discrete random variable X has the following probability distribution:
| \(x\) | \(-1\) | \(0\) | \(1\) | \(4\) |
| \(P(x)\) | \(0.2\) | \(0.5\) | \(α\) | \(0.1\) |
Determine the following:
1. \(α\)
2. \(P (0)\)
3. \(P (X > 0)\)
4. \(P (X ≥ 0)\)
5. The mean \(\mu\) of \(X\)
6. The variance of \(X\)
7. The standard deviation of \(X\)
Solution
1. Since all probabilities must add up to \(1\),\( α = 1 – (0.2 + 0.5 + 0.1) = 0.2\)
2. Referencing the table, \(P (0) = 0.5\)
3. From the table once again, \(P (X > 0) = P (1) + P (4) = 0.2 + 0.1 = 0.3\)
4. From the table, \(P (X ≥ 0) = P (0) + P (1) + P (4) = 0.5 + 0.2 + 0.1 = 0.8\)
5. Using the formula in the definition for mean \(\mu\):
\[\mu=E(x)=\sum x P(x)= ((-1)\cdot 0.2) + ((0)\cdot 0.5)+((1)\cdot 0.2)+((4)\cdot 0.1)=0.4 \]
6. Using the value of \(\mu\) obtained with the formula for variance:
\[\sigma^2 = \sum (x-\mu)^2P(x)=((-1-0.4)^2\cdot 0.2) + ((0-0.4)^2\cdot 0.5+ ((1-0.4)^2\cdot 0.2)+((4-0.4)^2\cdot 0.1)=1.84 \]
7. Using the calculated variance, we can then obtain the standard deviation using its formula as follows:
\[\sigma=\sqrt{\sum(x-\mu)^2P(x)}\]
\[\sigma= \sqrt{1.84}=1.3565\]
Discrete Random Variable - Key takeaways
- Discrete random variables are random variable that takes specified or finite values in an interval. Values may be countable and have a finite number of outcomes.
The types of discrete random variables are: Bernoulli, Multinomial, Binomial, Geometric, Hypergeometric, and Poisson.
A list of each potential value of a discrete random variable \(X\), along with the likelihood that \(X\) will take that value in one trial of the experiment, is the probability distribution of that discrete random variable \(X\).
The probability distribution for a binomial random variable is given by:
\[P(X=x)={\frac{n!}{(n-X)!}\cdot (X!)p^x q^{n-x}}\]
The probability distribution for a geometric random variable is given by:
\[P(X=x)=(1-p)^{x-1}p\]
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