# Discrete Probability Distributions

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Lesson Priority: VIP Knowledge

Objectives
• Understand what a probability distribution is and what it means
• Be able to differentiate discrete outcomes from continuous ones
• Know what properties are unique to discrete distributions
• Understand how discrete distributions may have a finite or infinite number of outcomes
Lesson Description

As we become more familiar with probability, it becomes necessary to work with specific types of probability distributions. This lesson introduces discrete distributions (separate countable outcomes), and specifies several properties that they must have to be legitimate. We'll also see both tabular and function forms of defining the potential outcomes and their respective probabilities.

Practice Problems

Practice problems and worksheet coming soon!

## Theoretical Distributions

With a firm understanding of what probability is, we can begin to study probability distributions. Up until now, we've worked with probabilities that were either pre-defined or success/failure based, such as the probability of flipping heads on a fair coin, or being told that the probability that Johnny gets an A in history this year is $75\%$. Probability distributions allow us to look at events with multiple outcomes.
Define: Probability DistributionA probability distribution is a list of outcomes for a random variable, and the associated probability of each outcome. It should accounts for all possible outcomes, so that the sum of all probabilities listed is $100\%$ or $1$.
Here's a quick example. Let's let the random variable $X$ represent the number of exams you have on a random day in school next year, assuming you never have more than $3$ on the same day. The probability distribution for $X$ could look something like this:$$\begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 55\% \\ \hline 1 & 30\% \\ \hline 2 & 10\% \\ \hline 3 & 5\% \\ \hline \end{array}$$Remember that when we talk about random variables », we are thinking about the unknown future, so the way we interpret this distribution is that on some randomly picked future school day, there is a $55\%$ chance that you will have zero exams that day, a $30\%$ chance that you will have one exam, $10\%$ chance for two exams, and $5\%$ chance for three exams. Once you know how many exams you'll have on that school day, we are no longer applying probability - at that point you would know which outcome happened.
You Should Know In probability, we commonly use capital letters to represent random variables, which are variables representing an unknown future outcome. When we describe possible outcomes in a distribution, we often use the same letter in lowercase to show the possibilities. Therefore, in the above illustrative example, $x$ represents various possible outcomes of the random variable $X$.
Probability distributions like this one are particularly useful in real-world probability applications (and yes, probability is highly utilized in business!) because it quantifies all possibilities and allows us to measure things like average expected outcome and how much variability exists in the unknown future.Probability distributions must follow some very basic rules so that the overall ideas of probability that you've learned up to this point are preserved. Namely:
• The probability of each outcome is between $0$ and $1$, where $0$ means the outcome is impossible and $1$ means the outcome is certain to happen
• The sum of all probabilities in a distribution must equal exactly $1$, so that the distribution contains every possibile outcome, since a sum to $1$ captures all event outcomes
We can see that the illustrative example above does indeed follow these two rules. We say that distributions that break these rules are invalid.

## Telling Apart Discrete from Continuous

We recently learned that random variables » can be either discrete or continuous. Discrete random variables are ones in which the outcomes are countable, while continuous random variables are ones for which it makes sense to have partial values for outcomes. Probability distributions exist for both cases, but this lesson focuses on discrete outcomes. After all, it is only the discrete outcomes that we can make a list for, since they are countable. If you are working with a continuous random variable, like time, distance, or sometimes money, do not try to apply the table format of distributions that we work with in this lesson.

Example 1Determine which of the following random variables is discrete and which is continuous.
• The random variable for the number of flowers in a randomly selected house's backyard
• The random variable for the number of coins a person can take in one grab out of a jar full of coins
• The random variable for the distance between two best friends at your school at any random time
• The random variable for amount of electricity a randomly selected house in your neighborhood used last month
$\blacktriangleright$ Let's quickly say what each of these is and why:
• The random variable for the number of flowers in a random house's backyard
• This is a discrete situation because the outcome is countable, even though the number probably has a large variance and high expected number in many cases.
• The random variable for the number of coins a person can take in one grab out of a jar full of coins
• Even though this is related to money (which is often considered continuous due to partial values), this situation measures the number of coins grabbed, which is a countable outcome and therefore discrete.
• The random variable for the distance between two best friends at your school at any random time
• Distance is almost always a continuous measure because we do not use individual countable units of measure but rather measure in whatever unit, such as feet, meters, or yards, and allow for partial values, e.g. $15.4$ meters.
• The random variable for amount of electricity a randomly selected house in your neighborhood used last month
• Electricity is not measured in a countable way like the number of beans in a jar is. Electricity is a continuous measure because we can have partial values (whether or not you know exactly what units we measure amounts of electricity in, hopefully this is a familiar concept to you).

## Finite Outcome Distributions

While this lesson focuses only on discrete situations, we can still have a finite or an infinite number of outcomes. Fortunately, having a finite number of outcomes is by far the most common case, which means we can jot down every outcome in table form alongside the probability that it happens. This is necessarily the case when we work with distributions that have non-number outcomes, such as we'll see in Example 2.Every valid probability distribution has the feature that the sum of all outcome probabilities is $1$. If this is not the case, then the distribution is either invalid or missing outcomes from its list. Make sure to double check that this is the case.Table form is very convenient for several types of common calculations. If a finite outcome discrete situation is described to us in words, we may need or want to turn it into tabular form.

Example 2James is not a consistent bowler, but believes that every time he bowls a frame, he is subject to a probability distribution such that he has a $10\%$ chance of bowling a strike, a $20\%$ chance of bowling a spare, and assumes the rest of his frames are not strikes or spares. Write the tabular form of this categorical discrete distribution.$\blacktriangleright$ Since James bowls a strike $10\%$ of the time and a spare $20\%$ of the time, the rest of his attempts that are not strikes or spares must happen $70\%$ of the time.$$\begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline \mathrm{Strike} & 10\% \\ \hline \mathrm{Spare} & 20\% \\ \hline \mathrm{Other} & 70\% \\ \hline \end{array}$$This is a categorical distribution because the outcomes are labels, not numbers, and therefore we cannot find quantities such as the "average" bowling result (more to come on that in the next lesson on expected values and variance »).

## Function-Defined Distributions

While we are most commonly given a distribution in table form or a word description, it is also possible to use function notation to describe distributions. We can get any individual outcome probability by plugging in the $x$ value. If there are few enough possibilities, we can also use the function to generate a table.
Define: Probability Mass FunctionA function $f(x)$ is called a probability mass function of random variable $X$ if $f(x)$ gives the probability of each input value $x$, and if the complete domain of possibilities and likelihoods form a valid probability distribution.

Example 3Use the given probability mass function$$f(x) = \frac{x^2 + 1}{35}$$to generate a probability distribution table for random variable $X$ whose only outcomes are $0$, $1$, $2$, $3$, and $4$.$\blacktriangleright$ Typically for a question like this, all we need to do is plug in each $x$ possible outcome. We should also quickly check that the distribution is valid.$$f(0) = \frac{1}{35}$$$$f(1) = \frac{2}{35}$$$$f(2) = \frac{5}{35}$$$$f(3) = \frac{10}{35}$$$$f(4) = \frac{17}{35}$$Therefore the resulting distribution is$$\begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & \frac{1}{35} \\ \hline 1 & \frac{2}{35} \\ \hline 2 & \frac{5}{35} \\ \hline 3 & \frac{10}{35} \\ \hline 4 & \frac{17}{35} \\ \hline \end{array}$$The five probabilities are all valid and sum to $1$, so this is a valid distribution.Note that with a denominator like $35$ it's best to leave the distribution likelihoods in fraction form. If we needed to we could get decimal approximations but exact answers are often more convenient.

## Infinite Outcome Distributions

While they are not nearly as common, it is likely that you will encounter some discrete distributions with an infinite list of possibilities. These distributions will always have a formula-based approach for calculating outcome probability, and typically have special names. One of the most common ones is called the Poisson Distribution » and works well for measuring outcomes that involve rates.We'll save the deep-dive of the Poisson distribution for its own lesson (if you even need to learn it - it's not on the AP Stats syllabus), but leave you with a quick example of what the formula and distribution may look like in context.

Example 4Suppose the probability that $x$ number of people walk into a particular small town bank in a given minute is given by the formula$$\mathrm{Pr}(X=x) = \frac{3^x e^{-3}}{x!}$$where the random variable $X$ represents the number of people that walk into this bank in a randomly selected minute during the bank's open hours (as you may learn one day, this means that $X$ follows a Poisson distribution with an average outcome of $3$).In table form, list the probability that $X$ is $0$, $1$, $2$, $3$, $4$, or more than $4$.$\blacktriangleright$ For a Poisson distributed random variable such as this example, there is no maximum for the number of people that could walk into the bank in that minute. We know it's unlikely that $50,000$ people will, but the technically it could happen and the probability formula allows us to plug in $50,000$ and get a probability for that outcome (though the probability is so close to zero it might as well be impossible).
You Should Know We won't prove this now, but if you add up the probabilities that $0$, $1$, $2$,... all the way to infinity number of people enter the bank in one minute, you will get exactly $1$. That's what makes this a valid probability distribution. However, after a certain large value of $x$, the probabilities for each further outcome are so small we can pretend it is literally zero. You probably won't be asked to prove this either since it requires infinite series which we study in second semester college Calculus.
For now, we were asked to simply write down six outcomes and the probabilities associated with these six outcomes.The probability that $X$ is $0$ can be found by plugging $0$ into the formula:$$\mathrm{Pr}(X=0) = \frac{3^0 e^{-3}}{0!} = \frac{1}{e^3} \approx 0.0498$$Each subsequent probability can be found by similarly plugging in:$$\mathrm{Pr}(X=1) = \frac{3^1 e^{-3}}{1!} = \frac{3}{e^3} \approx 0.1494$$$$\mathrm{Pr}(X=2) = \frac{3^2 e^{-3}}{2!} = \frac{9}{2e^3} \approx 0.2240$$$$\mathrm{Pr}(X=3) = \frac{3^3 e^{-3}}{3!} = \frac{27}{6e^3} \approx 0.2240$$$$\mathrm{Pr}(X=4) = \frac{3^4 e^{-3}}{4!} = \frac{81}{24e^3} \approx 0.1680$$Now, in order to find the probability that more than $4$ people walk into the bank, we must realize that all probabilities (up to infinity) sum to $1$, but that we have found the probability for each case up to and including $4$ - this means that the probability that more than $4$ people walk into the bank in a random minute is$$1 - \big[ \mathrm{Pr}(X=0, \, 1, \, 2, \, 3, \,\, \mathrm{or} \,\, 4) \big]$$$$\Rightarrow \mathrm{Pr}(X \gt 4) = 1 - \big[ 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 \big]$$$$=0.1848$$

## Probabilities of Ranges of Outcomes

In order to quickly calculate commonly requested quantities, mathematicians created two types of functions related to probability outcomes of discrete distributions - the PDF and the CDF.The PDF (Probability Density Function) describes the probability associated with a specific outcome, while the CDF (Cumulative Distribution Function) describes the probability of attaining a specific outcome or fewer. You probably don't need to memorize what the acronyms stand for, but you should have a strong understanding of each of these objects.
Define: The PDF and CDF of a DistributionThe Probability Density Function of a discrete distribution gives the probability that a specific outcome occurs. This can be stated for a specific value, e.g.$$\mathrm{PDF}(3) = 0.25$$or it can be described formulaically, e.g.$$\mathrm{PDF}(x) = {10 \choose x} (0.1)^x (0.9)^{10-x}$$The Cumulative Distribution Function of a discrete distribution gives the probability that a specific outcome or fewer will occur. Similarly to the PDF, this can be stated or described formulaically. For example, when we say$$\mathrm{CDF}(4)$$we are simply using shortcut notation to say$$\mathrm{PDF}(0) + \mathrm{PDF}(1) + \cdots + \mathrm{PDF}(4)$$It is the sum of all probabilities of outcomes that are less than or equal to the desired number.
You Should Know Some teachers absolutely insist on referring to the PDF as the PMF (probability mass function) for discrete distributions, and reserve the title PDF for continuous distributions. In my experience, both titles are fine, and people will know what you mean either way.
We should be thinking of PDF as a synonym to the notation $\mathrm{Pr}(X=x)$. It is just the probability of a certain outcome happening. The CDF however is the cumulative probability of all outcomes up to and including a value.

Example 5Given the following discrete probability distribution of the random variable $X$:$$\begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 5\% \\ \hline 1 & 25\% \\ \hline 2 & 20\% \\ \hline 3 & 25\% \\ \hline 4 & 15\% \\ \hline 5 & 10\% \\ \hline \end{array}$$Determine the following quantities:$$1) \,\, \mathrm{PDF}(2)$$$$2) \,\, \mathrm{CDF}(3)$$$$3) \,\, \mathrm{Pr}(X \gt 1)$$$\blacktriangleright$ The first two questions are straight definitions of these objects.The PDF of $2$ is asking what the probability that $X$ is $2$ is. From the table, this is simply $20\%$.The CDF of $3$ is asking what the probability that $X$ is $3$ or fewer is. This is merely the sum of the probabilities for the outcomes $0$, $1$, $2$, and $3$.$$\mathrm{CDF}(3) = 0.05 + 0.25 + 0.20 + 0.25$$$$= 0.75$$Finally, the question about what the probability of $X$ being greater than $1$ - this can be expressed in terms of the CDF, or done manually (line by line). For a table with only six rows, either approach is fine. However, if the table had $10$ or $20$ rows, the CDF approach is king:$$\mathrm{Pr}(X \gt 1) = 1 - \mathrm{CDF}(1)$$The idea here is that if the CDF of $1$ covers the probabilities of $X$ having the outcome of $0$ or $1$, and all probabilities sum to $100\%$ (or $1$), then $1 - \mathrm{CDF}(1)$ captures the probabilities of all outcomes that are larger than $1$. This is sometimes referred to as "the complement rule" (we'll look at this in more detail in its own section below).$$\mathrm{Pr}(X \gt 1) = 1 - \big[ \mathrm{Pr}(X=0) + \mathrm{Pr}(X=1) \big]$$$$\mathrm{Pr}(X \gt 1) = 1 - \big[ 0.05 + 0.25 \big]$$$$=0.7$$The manual or line-by-line approach would have been to literally add up the probabilities of all outcomes that are larger than $1$:$$\mathrm{Pr}(X \gt 1) = \mathrm{PDF}(2) + \mathrm{PDF}(3) + \mathrm{PDF}(4) + \mathrm{PDF}(5)$$$$= 0.20 + 0.25 + 0.15 + 0.10$$$$=0.7$$Same answer, and not much difference in terms of difficulty, but again, think about how much easier the first approach is if the table had dozens of rows.

Pro Tip Teachers and word problems like to leverage the CDF technique we saw at the end of the previous example. By requiring you to do something like $1 - \mathrm{CDF}(k)$, it forces you to better understand what the CDF is and how it can be used.

## The Complement Rule

When we have to work with table forms of distributions, or in other cases that require by-hand calculations, we can leverage a shortcut called "the complement rule of probability".Consider the following distribution for the random variable $X$:$$\begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 0.04 \\ \hline 1 & 0.07 \\ \hline 2 & 0.13 \\ \hline 3 & 0.16 \\ \hline 4 & 0.20 \\ \hline 5 & 0.15 \\ \hline 6 & 0.10 \\ \hline 7 & 0.06 \\ \hline 8 & 0.03 \\ \hline 9 & 0.01 \\ \hline 10 & 0.04 \\ \hline 11 & 0.10 \\ \hline 12 & 0.06 \\ \hline 13 & 0.03 \\ \hline 14 & 0.01 \\ \hline 15 & 0.04 \\ \hline \end{array}$$What if we were asked for the CDF of 13? It would be tremendously more difficult to get it by adding up the cumulative probabilities for outcomes zero through thirteen than it would if we remembered that all probabilities add to $1$, and so we can start with $1$ and subtract away the outcomes that we don't want. This shortcut is called the complement rule because in probability, the complement of any probability is $1$ minus that probability. If the probability represents the likelihood that the outcome will happen, the complement represents the probability that the outcome won't happen.$$\mathrm{CDF}(13) = 1 - \mathrm{Pr}(X=14) - \mathrm{Pr}(X=15)$$$$=1 - 0.01 - 0.04 = 0.95$$

## Mr. Math Makes It Mean

Conditional ProbabilityConditional probability » (sometimes called dependent probability) is among the more confusing concepts in probability. We'll cover it in more detail a few lessons from now, but it's possible you may have already seen it in your classwork by the time you are beginning to learn the ins and outs of discrete distributions.In short, conditional probability requires a special formula to adjust for questions in which we know some bonus information about an outcome. If we know event $B$ happened, the probability that event $A$ happened may not be the same as if we didn't know that event $B$ happened. The formula is$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}$$The biggest challenge in using conditional probability with any random discrete distribution we may be working with is simply putting all the knowledge and details together correctly. Make sure to look at the dedicated lesson » if you are new to conditional probability, and feel free to skip this section for the time being if your class hasn't got there yet (though learning by example in this lesson will be very helpful!).

Example 6A special six-sided die is manufactured unbalanced (sometimes referred to as a "loaded" die) such that the probabilities of its six outcomes are given by the following table:$$\begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 1 & 5\% \\ \hline 2 & 5\% \\ \hline 3 & 10\% \\ \hline 4 & 10\% \\ \hline 5 & 30\% \\ \hline 6 & 40\% \\ \hline \end{array}$$Let $X$ be the random variable for the outcome of a random roll with this unfair die. Answer the following questions about $X$:$1)$ $\mathrm{Pr}(X=2 | X \leq 4)$$2) \mathrm{Pr}(X=4 | X=\mathrm{even})$$3)$ \mathrm{Pr}(X=\mathrm{even} \, | X \ge 3)$$\blacktriangleright All three of these questions are conditional, so let's jot down the general formula for a conditional probability calculation:$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}$$Where A is the event we want the probability of, and B is the event that we assume is definite to occur. 1) \;\;\;\; \mathrm{Pr}(X=2 | X \leq 4)For the first question, we need the probability that X is less than or equal to 4 (the \mathrm{Pr}(B) term):$$\mathrm{Pr}(X \leq 4) = 0.05 + 0.05 + 0.10 + 0.10 = 0.30$$We also need the probability that X is 2 and X is less than or equal to 4 (the \mathrm{Pr}(A \cap B) term). Because X is already less than or equal to 4 when X is 2, the probability that X is 2 and X is less than or equal to 4 is equal to the probability that X is 2 (another way to think about this is that X=2 is the only outcome that satisfies both requirements).$$\mathrm{Pr}(X=2 \cap X \leq 4) = \mathrm{Pr}(X=2) = 0.05$$Finally, plug the pieces into the conditional formula.$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}\mathrm{Pr}(X=2 | X \leq 4) = \frac{0.05}{0.30}=\frac{1}{6}\blacksquare$$2) \;\;\;\; \mathrm{Pr}(X=4 | X=\mathrm{even})We'll answer the second question in a similar way.$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}$$In this case, A is the event that X is 4, and B is the event that X is even.The probability that both things happen is 10\%, because the only way for X to be both 4 and even is if X is 4.The probability that X is even is the sum of 5\%, 10\%, and 40\%, since those are the respective probabilities of rolling a 2, 4, or 6.Putting these pieces together:$$\mathrm{Pr}(A | B) = \frac{0.10}{0.55} = \frac{2}{11}\approx 0.1818\blacksquare$$3) \;\;\;\; \mathrm{Pr}(X=\mathrm{even} \, | X \ge 3)Yet another conditional probability, let's start by labelling what we want:$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}$$For this question, we'll say A is the event that we roll an even number, and B is the event that we roll at least a 3.The probability that both things happen includes the outcomes of 4 and 6, since those are the only even numbers that are also big enough to be at least 3.$$\mathrm{Pr}(X=\mathrm{even}) = 0.10 + 0.40 = 0.50$$The probability that X is at least 3 is the sum of the outcomes 3 through 6.$$\mathrm{Pr}(X \ge 3) = 0.10 + 0.10 + 0.30 + 0.40 = 0.90$$Or, as an alternative approach, the probability that X is at least 3 can be found using the complement property: 1 - \mathrm{Pr}(X \lt 3):$$\mathrm{Pr}(X \ge 3) = 1 - \big( \mathrm{Pr}(X = 1) + \mathrm{Pr}(X = 2) \big) = 1 - 0.05 - 0.05 = 0.90$$As always, sometimes one method or the other is substantially less work, but here both are about the same.Putting this information together, we have$$\mathrm{Pr}(A | B) = \frac{0.50}{0.90} = \frac{5}{9}\approx 0.5556\blacksquare$$Table BuildingWord problems in which you must write down your own probability table are often difficult only due to how closely we need to read and interpret the wording. Make sure you are especially careful to look for things like "at least" versus "more than", which have two slightly different interpretations.Sometimes it's just a matter of tracking the details and writing the table such that probability rules are kept intact. In other cases, such as Example 7, you need to use both probability rules and just a bit of algebra representation to jot down the correct table. Example 7Henry's dog Max barks when food is scooped into Max's dog dish. Max doesn't always bark, but he never barks more than 3 times when he hears the scoop of food being dumped in his dish. The probability that he barks twice is twice as high as the probability he barks once, and the probability he barks three times is half as much as the probability he barks once. There is a 30\% probability that he doesn't bark at all. Calculate the probability associated with each possible outcome for the number of times Max barks at dinner time, and then describe the distribution in table form.\blacktriangleright Let's begin by calculating the probabilities of each outcome. Let p be the probability that Max barks once. Based on the wording of the problem, this means that the probability Max barks twice is 2p, and the probability Max barks three times is p/2. X is the random variable for the number of times Max barks at any particular meal time.If we like, we can get the table structure written at this point, before we move on to solving for p. In fact some students may prefer to do this, since it allows us to concisely see the universe of possibilities.$$ \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 30\% \\ \hline 1 & p \\ \hline 2 & 2p \\ \hline 3 & p/2 \\ \hline \end{array} $$We can solve for p at this point because we know that the sum of all possibile outcome probabilities must be 1, so that 100\% of the time, one of these four outcomes happens (based on our assumptions given to us in the problem, namely that Max barks between 0 and 3 times when being fed, and never any more than that).$$0.3 + p + 2p + p/2 = 10.3 + \frac{7p}{2} = 1\frac{7p}{2} = 0.7\Rightarrow p = 0.2$$Plugging in for p, this means our distribution is actually$$ \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 30\% \\ \hline 1 & 20\% \\ \hline 2 & 40\% \\ \hline 3 & 10\% \\ \hline \end{array} $$and we can verify that the probabilities both add to 1 and follow the rules we were told that the follow. ## Put It To The Test Example 8For each of the following random variables, determine if it is continuous or discrete.1) The amount of soup served in a restaurant on any given night2) The number of pennies in a coin jar3) The chair count at a randomly selected wedding hall4) The amount of wood needed to build a house Show solution 1) The amount of soup served in a restaurant on any given nightThere isn't likely an easy way to measure soup in a countable way, so this is almost definitely a continuous random variable (though I suppose if described properly it's possible to be discrete, like selling it only in one pint containers)2) The number of pennies in a coin jarThe number of pennies is definitely countable, even if there are a lot of them (no partial values make sense in this case). This is a discrete random variable.3) The chair count at a randomly selected wedding hallOnce again it doesn't seem to make sense to have partial values here, since a chair is a whole object. This is a discrete random variable.4) The amount of wood needed to build a houseIt's possible to describe wood in discrete countable ways, such as a number of two by fours or a number of pieces of plywood, so perhaps the jury is out on this one. However I think most people would describe this in weight, such as pounds or kilograms, and if so it would be a continuous measure. Example 9Which of the following distributions is valid?$$\begin{align} \mathrm{A)}\;\;\;\;\;\;\;\; & \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline -10 & 20\% \\ \hline -5 & 20\% \\ \hline 0 & 30\% \\ \hline 5 & 10\% \\ \hline 10 & 10\% \\ \hline 25 & 10\% \\ \hline \end{array} \\ \mathrm{B)}\;\;\;\;\;\;\;\; & \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & -30\% \\ \hline 1 & 50\% \\ \hline 2 & 40\% \\ \hline 3 & 10\% \\ \hline 4 & 30\% \\ \hline \end{array} \\ \mathrm{C)}\;\;\;\;\;\;\;\; & \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 30\% \\ \hline 1 & 20\% \\ \hline 2 & 40\% \\ \hline 3 & 20\% \\ \hline \end{array} \end{align} $$Show solution \blacktriangleright Only distribution A) is proper. B) does have the "sum to 1" rule satisfied, but it also contains outcomes with negative probability. C) is invalid because the sum of its outcome probabilities are greater than 1. Example 10You are given the following distribution for X, the number of hours it takes for a randomly selected college student to travel home at the end of the semester, assuming a maximum of ten.$$ \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 0.04 \\ \hline 1 & 0.07 \\ \hline 2 & 0.13 \\ \hline 3 & 0.16 \\ \hline 4 & 0.20 \\ \hline 5 & 0.15 \\ \hline 6 & 0.10 \\ \hline 7 & 0.06 \\ \hline 8 & 0.03 \\ \hline 9 & 0.01 \\ \hline 10 & 0.04 \\ \hline \end{array} $$Find the following quantities:1) \mathrm{PDF}(3)$$2) $\mathrm{CDF}(2)$$3) \mathrm{CDF}(9)$$4)$ $\mathrm{Pr}(X$ is divisible by $3)$$5) \mathrm{Pr}(X \gt 3) Show solution 1) \mathrm{PDF}(3):All we need to answer this question is to remember what PDF means. This is a fancy way of asking \mathrm{Pr}(X=3), which is right in the table.$$\mathrm{PDF}(3) = 0.16\blacksquare$$2) \mathrm{CDF}(2):Here we must remember that the CDF of 2 is the cumulative sum of all probabilities of outcomes 2 and under. For this question that means we need to include the outcomes 0, 1, and 2.$$\mathrm{CDF}(2) = 0.04 + 0.07 + 0.13 = 0.24\blacksquare$$3) \mathrm{CDF}(9):For this question is pays to pay attention to the distribution. The CDF of 9 is the cumulative sum of probabilities for outcomes 0 to 9, but we don't have to add ten things together. Instead, since the distribution only goes up to 10, we can use the complement rule.$$\mathrm{CDF}(9) = 1 - \mathrm{Pr}(X=10) = 0.96\blacksquare$$4) \mathrm{Pr}(X is divisible by 3):It's up to us to remember what divisibility is all about and pick out the outcomes we want. Add up the probabilities of the outcomes that satisfy this requirement:$$\mathrm{Pr}(X \mod 3 \equiv 0) = \mathrm{Pr}(X=3, \; 6, \; \mathrm{or} \; 9)= 0.16 + 0.10 + 0.01 = 0.27\blacksquare$$5) \mathrm{Pr}(X \gt 3):This is also a complement rule question. It's much faster to express this question using a CDF:$$\mathrm{Pr}(X \gt 3) = 1 - \mathrm{CDF}(3)= 1 - \big( 0.04 + 0.07 + 0.13 + 0.16 \big) = 1- 0.4 = 0.6\blacksquare$$Example 11You are given the following discrete probability distribution for the random variable X, the number of hours an AP calculus student spends each week on homework:$$ \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 0.05 \\ \hline 1 & 0.05 \\ \hline 2 & 0.10 \\ \hline 3 & 0.20 \\ \hline 4 & 0.20 \\ \hline 5 & 0.20 \\ \hline 6 & 0.10 \\ \hline 7 & 0.05 \\ \hline 8+ & 0.05 \\ \hline \end{array} $$Calculate the following probabilities:1) \mathrm{Pr}(X \ge 1 | X \lt 4)$$2)$ $\mathrm{Pr}(X \gt 5 | X \gt 2)$$3) \mathrm{Pr}(X = 3 | (X \ne 5 \cap X \ge 2) Show solution 1) \mathrm{Pr}(X=0 | X \lt 4):As always, when working with the conditional probability formula, we should start by jotting down the generic formula and labelling the events:$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}$$Let's say A is the event that the student studies 1 or more hours, and B is the event that the student studies less than 4 hours.The formula requires us to know the probability that events A and B both happen, as well as the probably that event B happens overall.A and B both happen when the student studies for 1, 2, or 3 hours, and so$$\mathrm{Pr}(A \cap B) = 0.05 + 0.10 + 0.20 = 0.35$$Event B happens overall when the student studies for less than 4 hours, which could be the outcomes 0, 1, 2, or 3.$$\mathrm{Pr}(B) = 0.05 + 0.05 + 0.10 + 0.20 = 0.40$$Therefore, using the conditional probability formula, we have$$\mathrm{Pr}(X \ge 1 | X \lt 4) = \frac{0.35}{0.40} = \frac{7}{8}=0.875\blacksquare$$2) \mathrm{Pr}(X \gt 5 | X \gt 2):Once again, label events A and B in the conditional formula, and calculate each piece.A is the event that X is 6, 7, or 8+.B is the event that X is 3, 4, 5, 6, 7, or 8+.Therefore A and B can only happen together if X is 6, 7, or 8+.$$\mathrm{Pr}(A \cap B) = 0.10 + 0.05 + 0.05 = 0.20\mathrm{Pr}(B) = 0.20 + 0.20 + 0.20 + 0.10 + 0.05 + 0.05 = 0.80$$Therefore,$$\mathrm{Pr}(X \gt 5 | X \gt 2) = \frac{0.20}{0.80} = \frac{1}{4}=0.25\blacksquare$$3) \mathrm{Pr}(X = 3 | (X \ne 5 \cap X \ge 2):Follow along with the same formula we used for the last two questions. The only outcome that satisfies both conditions is when X is 3, and the probability of the given event is the outcome probabilities for X being 2, 3, 4, 6, 7, and 8+.$$\mathrm{Pr}(X = 3 | (X \ne 5 \cap X \ge 2) = \frac{0.20}{0.70} = \frac{2}{7}\approx 0.2857\blacksquare$$Example 12In any month, your car might require a maximum of three repairs. If the probability of needing two or three repairs is each 5\%, and the probability of needing one repair is 25\%, write a probability distribution in table form for the outcomes of needed car repairs in any random future month. Show solution \blacktriangleright The probability rule we need to make sure we've thought about is the "sum to 1" rule. Do all the outcomes' probabilities add to 1? They do if you include the implied outcomes of zero needed repairs, and use the sum to 1 rule to figure out the probability that no repairs are needed.This is not guesswork - the problem said that the car could require up to three repairs, so we do not allow the possibility for this situation that we could need four or more. We also know logically that a car cannot need negative repairs, so when we notice that the given probabilities do not sum to 1, it is our job to realize that the remaining probability belongs to the case where the car does not need any repair.$$ \begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 65\% \\ \hline 1 & 25\% \\ \hline 2 & 5\% \\ \hline 3 & 5\% \\ \hline \end{array}$\$

Lesson Takeaways
• Understand how distributions describe the likelihood of outcomes in any situation
• Know the core characteristics that must be true of any probability distribution
• Continue to develop an understanding of the difference between discrete and continuous situations
• Be able to write down your own table of outcomes and probabilities based on a word problem description
• Know when discrete outcomes need to include or not include zero, and when it makes sense for outcomes to theoretically be infinitely large even though such outcomes are unlikely to happen (see the bank example)
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