The Binomial Distribution

Lesson Features »

Lesson Priority: VIP Knowledge

  • Know the specific circumstances that define a binomial distribution
  • Understand what probability the binomial distribution measures
  • Know and memorize the formulas for the mean and the standard deviation of a binomial distribution
  • Master the difference between the "pdf" and the "cdf"
  • Be able to get fast answers from your TI-84 Calculator for binomial problems
Lesson Description

Due to its simplicity and far-reaching applicability, the Binomial Distribution is one of the most common discrete distributions. It quantifies how likely it is to have a certain number of successes in $n$ attempts, given a fixed per-trial success probability. E.g. this distribution can tell you how likely it is that you got 10 out of 20 SAT questions correct if you were guessing blindly. This lesson will introduce both the concept of the binomial distribution as well as the formulas you will be expected to retain for quizzes and tests on this distribution.

Practice Problems

Practice problems and worksheet coming soon!


Measuring Success

The Binomial Distribution measures the probability that, out of $n$ identical attempts at a success / failure objective, $k$ of those attempts will be successes.In order to use this distribution correctly, we need a few things. First, we need to understand when this concept does and does not apply. Second, we need to understand what we're measuring and what the correct interpretation of the probabilities and expected values is. Finally, we need to be able to get these results practically, which depends entirely on how much your teacher allows you to depend on Calculator Magic.

When Can I Binomial?

As we said, the probabilities we calculate with the Binomial Distribution are for success / failure situations. Specifically, Binomial random variables measure the number of successes only for situations involving repetitive trials of the same objective. Because it is the same repetitive objective, the Binomial Distribution formulas require that the probability of each trial being a success must be the same. This is a very important requirement that trick questions will test you on!We also need to be able to assume that successive trials do not affect one another - in other words, that they are independent.Here are some examples of situations that we can measure with the Binomial Distribution:
  • The probability for the number of times we get an even roll when we roll a fair die 10 times
  • The probability for the number of correct multiple choice SAT questions a zoo animal answers correctly out of 20
  • The probability for how many times a particular artist's song shows up during an hour of internet radio
In each of these cases, the repetitive action that could turn out to be success or failure is 1) repetitively the same action 2) the same probability of success for each attempt and 3) independent in each attempt.However, we need to state that we believe the underlying conditions are truly met - particularly independence. For example, in the SAT example, it would need to be true that each question had the same number of answer choices, and we would also have to believe that the animal was choosing truly randomly. In the Internet radio example, we would need to believe that the playlist is shuffled randomly, and that each time a new song comes up, the probability that a song by that artist comes up next is always the same regardless of prior song plays.This isn't usually a big deal. Questions and situations will often state something like "assume independence", or we just state that we assume it exists. It's just something that needs to be said.Here are some examples of situations that have outcomes that cannot be measured by the Binomial Distribution:
  • The probability for how many times out of ten dice rolls that you roll a "four", where each time you use a different die, using some die that are six-sided and some that have more sides
    • The fact that each die roll is using a different die, and that each die may or may not be a traditional six-sided die, ruins the required assumption that each trial has the same success probability
  • The probability for hitting a certain number of bulls-eyes thrown by a dart thrower who is throwing 100 darts in a row
    • This isn't a situation where random chance is the only factor, and it is probably not true that this dart thrower has the same chance of success of hitting a bulls-eye on each attempt - reasonably, successive throwing may tire him out, making later throws less likely to succeed.
You Should Know
Sometimes it's up to us to judge the suitability of a situation as to whether the binomial distribution is fair to use, and sometimes we are just told to go with it.If we are told to assume consistency and independence then we should do so. For example, we could be told that the dart thrower has a perfect 75% success per throw probability and that we can assume all 100 of his throws are independent. It's just worth pointing out that if it's up to us to judge independence, we would be justified in thinking it probably doesn't exist in that situation.However, in the example where different sided dice are being used in different trials, there is no way that can be binomial because we'll never have a consistent success probability.

What Are We Quantifying?

In general, a probability distribution gives you a probability associated with a specific outcome. It is both helpful and necessary to have a strong understanding of what you're looking at when you get a result from a binomial situation.In any binomial situation you're examining, there are $n$ identical trials being done repetitively, with each trial having a probability of $p$ of being successful. The probabilities we get as answers tell us the likelihood of $k$ number of trials being successful, where $k$ is a number smaller than $n$.Follow along with the descriptions, probabilities, and notations below to put this concept into context.
Example 2You are blindly grabbing single M&M's from a bag and putting them back. You have been told that the probability of drawing a blue M&M is $15\%$. You would like to know the probability that you draw and put back $10$ individual candies, $3$ of which are blue.$\blacktriangleright$ First, we can validate that this is a binomial random situation. Each draw attempt has a fixed success probability of $15\%$ (keeping in mind that "success" in this situation means you picked a blue one), and each draw is independent because we are drawing from the same pool of candy (putting back each candy after it is drawn, which isn't very sanitary but is necessary to maintain independence here).Now, we say $X$ is the random variable that represents the number of blue candies that will be drawn. We do not know what the outcome will be before we conduct this experiment, but we can calculate the probability that $X$ is $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, or $10$, since those are the only possible numbers of blue candies we can draw when we make ten draws.The way to notate the probability that $X$, the number of blue candies that will be drawn is $3$, is:$$P(X=3)$$or$$\mathrm{Pr}(X=3)$$(both are used interchangeably by books, teachers, and profs based on their preference)And, when we calculate the probability of this happening to be approximately $13\%$ (you'll see how soon), we can say the following correct sentence to interpret this result:The probability that out of $10$ draws, $3$ of them are blue is approximately $13\%$, if the probability of each draw being blue is $15\%$.Now, let's learn how to find these probabilities.
Pro Tip
A binomial variable with $n=1$ is called a Bernoulli random variable. Bernoulli variables are useful in certain advanced probability applications because of their variance, but while you might hear this name in your class, they probably won't be very useful to you unless you take an advanced college probability course, or study actuarial science.

The Binomial Formula

The fortunate among you, those who know their class is going to 100% rely on a TI-84 or similar calculator, can likely read this section as bonus knowledge. While you do need some sort of calculation to carry out the formula, the formula is quickly becoming "the old" way of doing this, since TI-84 type calculators do the formula for you.
Define: Binomial Outcome ProbabilityThe probability of observing $k$ successes out of $n$ independent and identical attempts, each with probability of success $p$, is given by the formula$$\mathrm{Pr}(X=k) = {n \choose k} \,\, p^k \, (1-p)^{n-k}$$
You might wonder why this seemingly simple idea requires so thick a formula. In short, the first part of the formula (the nCr combinations) accounts for the different order in which the successes may occur, while the $p$ and $1-p$ pieces calculate the probabilities of having the specified number of successes and failures. If you have $n$ trials and $k$ successes, then you have $n-k$ failures.
You Should Know
Many teachers and profs favor short-handing the compliment probability $1-p$ with the letter $q$. It is simply a matter of abbreviation. The idea is that, in a prob and stats course, we reference $1-p$ so much that it's nice to have a symbol to represent it. It is also very prevalent to do this in actuarial science when quantifying contingent probabilities.Many textbooks do this as well. If you replace $1-p$ with $q$ in this formula, you'll see the formula that appears in many books:$${n \choose k} \,\, p^k \, q^{n-k}$$

Binomial Mean

We need to know what mean or expected value that a binomial distribution has, which is something we'll often want to know with any probability distribution. Unlike a "straight average" which you probably know how to do with some given values, the mean of a binomial distribution represents the average outcome we would expect if the entire experiment was conducted millions of times.For context, in our $15\%$ probability $10$-draw M&M example, we found that the probability of $X$ (the number of blue M&M's that were about to be drawn) being $3$ was about $13\%$. We could have done this for any of the other possible outcomes, $0$ through $10$, but that still doesn't answer the question - what would happen if we did this entire $10$-draw experiment millions of billions of times? What is the "average" outcome?There are two ways to proceed. In an earlier lesson », we learned that for any probability distribution, the mean, also known as the "expected value", can be obtained by adding the products of the values and the probabilities. For example, here is the full probability distribution for the blue M&M example.$$\begin{array}{|c|c|} \hline x & \mathrm{Pr}(X=x) \\ \hline 0 & 0.197\\ \hline 1 & 0.347\\ \hline 2 & 0.276\\ \hline 3 & 0.130\\ \hline 4 & 0.040\\ \hline 5 & 0.009\\ \hline 6 & 0.001\\ \hline 7 & 0.000\\ \hline 8 & 0.000\\ \hline 9 & 0.000\\ \hline 10 & 0.000\\ \hline \end{array}$$Probabilities are rounded to the nearest $0.001$, so while the likelihood of $7$ or more blue M&M's rounds to $0$, it is not actually equal to $0$ (but these outcomes are very, very close to zero because they are very, very unlikely to happen). The probabilities sum to $1$ as expected, since a probability of $1$ encompasses all possibilities.You can use this table to get the expected value of this distribution by adding the products of the values and their corresponding probabilities.$$\mu = (0)\cdot(0.197) + (1)\cdot(0.347)$$$$ + (2)\cdot(0.276) + \dots + (9)\cdot(0.000) + (10)\cdot(0.000)$$$$\approx 1.500$$Alternatively, there is a one-shot quick formula that will give you the expected value of a binomial distribution. It's a common theme in probability that "named" distributions, ones that have a name and formulaic probability approach, typically also have formulaic expected values.
Binomial Expected ValueThe expected value of a random variable $X$ which follows a binomial distribution with $n$ trials with a $p$ probability of success on each trial has an expected value of$$\mathrm{E}(X) = np$$Sometimes this is notated using the lowercase Greek letter mu ($\mu = np$).
For this situation, the expected value is $10 \cdot 0.15$, or 1.5. This matches our tedious by-hand calculation.
You Should Know
As we discussed in the prior lesson on general discrete probability distributions », the expectation of a distribution is like an average. It doesn't mean that you can necessarily obtain the expectation on any single outcome, but rather, it tells us what we would expected the experiment result to be on average, if we did the whole experiment millions of times.In this M&M example, we expect that on average, if we draw $10$ times, record the number of successful blue colored draws, and then repeat that $10$-draw process millions of times, that on average we will have drawn $1.5$ blue M&M's for every $10$-draw experiment. This happens to be a non-integer, so it's clear that while this is the average expectation, any single $10$-draw experiment could never result in having drawn $1.5$ blue candies, since we can only draw whole numbers of candies.

Binomial Variance and SD

The discussion about the variance and standard deviation of this distribution is nearly identical to the one we just had about expected value, in that there is a way to do it "the long way" using generic approaches we learned for probability distributions that work for the binomial distribution because they work for any distribution.However, the variance of a binomial random variable has a simpler formula, and we are expected to know it.
Binomial Variance and Standard DeviationThe variance of a random variable $X$ which follows a binomial distribution with $n$ trials with a $p$ probability of success on each trial has a variance of$$\mathrm{Var}(X) = npq$$where $q = 1-p$ is the complement probability of $p$.Sometimes this is notated using the lowercase Greek letter sigma, squared ($\sigma^2 = npq$).The standard deviation is the square root of the variance, as it is for any random variable.$$\mathrm{SD}(X) = \sqrt{npq}$$Using Greek letter notation, this would say $\sigma = \sqrt{npq}$.

PDF Versus CDF

Recall from the introductory lesson on Discrete Probability Distributions » that we work with two types of probability results: 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 associated with the outcomes that are less than or equal to a certain amount of successes.Looking back at our M&M's situation, we would say that PDF($6$) represents the probability that we draw six blue candies, while CDF($6$) represents the probability that we draw six or fewer candies.This distinction is plain enough, but the challenge will come from the wording of questions. It may seem like a cheap shot, but textbooks and teachers are infamous for tricky questions that can have slightly different meaning depending on changing a single word.It is worth a reminder that there is no formula for a discrete CDF. Beside using a calculator, the only thing you can do is add up individual PDF results. For example, CDF($6$) = PDF($0$) + PDF($1$) + PDF($2$) + PDF($3$) + PDF($4$) + PDF($5$) + PDF($6$). Fortunately, we will almost always use a calculator. Additionally, there is a complement trick for computing CDFs that are on the high end of the range, which we will see below.We'll see how to use PDF and CDF to answer the right question, but first let's know our way around the calculator.

Using Your Calculator

Using the TI-83 / 84 calculators, we can get instant probabilities for both PDF and CDF situations.First, choose DIST (by pressing 2nd $\longrightarrow$ VARS)Select binompdf, which is option "0" or "A" depending on your model. Also depending on your model, you will either be brought to a screen like thisor this, if your calculator software is older:The first case has a nice interface - just literally tell it what your $n$, $p$, and desired $x$ is. For example, for a binomial situation with $7$ trials with each trial's success probability equal to $20\%$, then $n$ is $7$ and $p$ is $0.2$. The desired $x$ is whatever probability you're asking for. For example, if you want to know the probability that $3$ trials are successful in this situation, $x$ is $3$.If you have older software and you have to enter the numbers yourself, you have to remember 1) the order to enter them ($n$ then $p$ then $x$), and 2) to use the comma button    ,    to separate each of the three numbers. When you're done, using either method, you should see something like this on your screen:Hit the enter button, and viola.The process for getting CDF probabilities from your TI is nearly identical. The only difference is you choose the option for "binomcdf" instead of "binompdf".

Answering the Right Question

The last thing you need to start answering quiz questions like a champ is to understand what question you're being asked. A good first step is to identify whether the probability they're asking for is a single outcome or a range of outcomes. Then, we should set up the appropriate PDF or CDF expression. The rest is done on your calculator (hopefully).Here is the binomial situation we will use to work through some clarifying examples.
Example 3A fair ten-sided die (a D-10 for the game nerds in the room) labeled $1$ to $10$ is rolled eight times. If the die roll is a $6$ or lower, the roller "wins" that roll. Let $X$ be the random variable that represents the number of wins that are achieved in this situation.Before we answer any questions about the unknown outcome of this situation, we should make sure it is a legitimate binomial situation, and if so, write down the $n$ and $p$. Repetitive dice rolls can be assumed to be independent from one another any time we are using a fair die, and each time we roll the probability is always the same because in between rolls we are not changing the rules of what defines a win or loss. This is a legitimate binomial distribution with $n=8$ and $p=0.6$ ($p$ is $0.6$ because $6/10$ dice roll outcomes will make that roll a "win"). A player may win as many as eight times and as few as zero.Question 1:What is the probability that the roller wins four times?$\blacktriangleright$ This is a PDF question, because PDF probabilities give the likelihood of one specific outcome.$$\mathrm{Pr}(X=4) = \mathrm{PDF}(4)$$Now that you have written down the appropriate PDF or CDF statement, use your calculator.The probability that a player rolls eight times and wins on exactly four rolls about $0.2322$.Question 2:What is the probability that the roller wins fewer than three times?$\blacktriangleright$ This is a CDF question, but we must take care with setting it up. The question wants us to find a probability for a range that uses a strict inequality, but our CDF calculations use inclusive inequalities. The best defense to being careful is to continue to set up PDF and CDF expressions on paper before we reach for our calculator.$$\mathrm{Pr}(X \lt 3) = \mathrm{CDF}(2)$$The CDF of $2$ will include the sum of the probabilities of achieving $0$, $1$, or $2$ wins. This is exactly what we want to include, to capture the described range of desired outcomes (fewer than three means we do not include three).Having answered the right question, we need only ask our calculator for the answer.$$\approx 0.0498$$Question 3:What is the probability that the roller wins between 3 and 7 times?$\blacktriangleright$ Any time we are asked for a range with a start and a finish, we can set it up as a difference of CDF probabilities rather than add up individual outcome PDF probabilities. The best way to think about the set up is to consider what you do and don't want to include, remembering that a CDF is an inclusive "less than or equal to" expression.For this example, we want to include the probabilities of winning $3$, $4$, $5$, $6$, or $7$ times. Again, we don't want to add up each probability separately. We would get the right answer but it is five separate calculations. Instead, start with $\mathrm{CDF}(7)$ which includes $7$ and under, and then subtract $\mathrm{CDF}(2)$ which includes $2$ and under.$$\mathrm{Pr}(3 \leq X \leq 7) = \mathrm{CDF}(7) - \mathrm{CDF}(2)$$Repeating the reason in detail for clarification, this is because$$\mathrm{CDF}(7) = \mathrm{Pr}(X = 0) + \mathrm{Pr}(X = 1) + \cdots + \mathrm{Pr}(X = 7)$$and$$\mathrm{CDF}(2) = \mathrm{Pr}(X = 0) + \mathrm{Pr}(X = 1) + \mathrm{Pr}(X = 2)$$so $\mathrm{CDF}(7) - \mathrm{CDF}(2)$ is equal to$$\Big[ \mathrm{Pr}(X = 0) + \mathrm{Pr}(X = 1) + \cdots + \mathrm{Pr}(X = 7) \Big] - \Big[ \mathrm{Pr}(X = 0) + \mathrm{Pr}(X = 1) + \mathrm{Pr}(X = 2) \Big]$$$$ = \mathrm{Pr}(X = 3) + \mathrm{Pr}(X = 4) + \cdots + \mathrm{Pr}(X = 7)$$which is exactly what we seek.Finally, we can quickly use a calculator to get each value, and subtract them. From the TI:$$\mathrm{CDF}(7) = 0.9832$$$$\mathrm{CDF}(2) = 0.0498$$Therefore, $\mathrm{CDF}(7) - \mathrm{CDF}(2) \approx 0.9334$.Question 4:What is the probability that the roller wins at least once?We can do this one with two perspectives. First, we could take the range perspective we used for Question 3. At least once means we seek$$\mathrm{Pr}( 1 \le X \le 8)$$which, using our CDF subtraction logic we just discussed, can be expressed as$$\mathrm{Pr}( 1 \le X \le 8) = \mathrm{CDF}(8) - \mathrm{CDF}(0)$$Plugging these requests into your calculator will give you the answer. However, this is a good time to understand a useful shortcut that I refer to as "the complement rule".In short, $\mathrm{CDF}(8)$ is $1$ (try to work out for yourself why that is), and $\mathrm{CDF}(0)$ is equal to $\mathrm{PDF}(0)$, since $0$ is the smallest number of wins you can have. This means that the probability we seek can be written as $\mathrm{Pr}(X=$ at least one)$=$ $1-\mathrm{Pr}(X = 0)$. In words, we say that the probability of at least one win is the probability complement to the outcome of winning zero times. If the probability you win zero times is $0.001$, then the probability you don't win zero times is $1-0.001$. And not winning zero times is a synonym for winning at least once.For now, just know that it's helpful to keep the complement rule in mind. It will always be true that the probability of something not happening is $1$ minus the probability that it does happen, and sometimes that fact is very convenient in a problem.$$\mathrm{Pr}( X \ge 1) = 1 - \mathrm{Pr}(X = 0) = 0.999$$
Pay close attention to the wording of any probability question that asks for specific outcome probabilities. The words "more than five" do not have the same meaning as the words "at least five" - one implies including $5$ as a possibility while the other doesn't. It's up to us to pay attention and keep it straight.
PDF and CDF SummaryIn summary, you should understand what PDF and CDF are, and be ready to use them for applied problems. The real difficulty is sorting through the detail. The key fact to remember is that a CDF expression includes the endpoint and everything below it. The real challenge is to correctly set up these problems, making sure you are including outcomes that you want and excluding outcomes that you don't.
Pro Tip
At the end of the day, just realize that each question can ask anything, and that the real challenge to these questions is to correctly set up the PDF or CDF expression that matches the word description.

Mr. Math Makes It Mean

Once you've mastered the ins and outs of binomial situations and probability calculations, teachers will often ask questions that implement conditional probability rules.Recall that the dependent probability formula is$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}$$Let's see an example of how this can be implemented in a binomial situation.
Example 4Jennifer is playing a spinner game, where $1/4$ of the spinner area is blue, and landing on blue is a "win" while landing anywhere else is a "lose". She plays this game five times in a row.What is the probability that Jennifer wins twice, given that she wins at least once?$\blacktriangleright$ First, let's identify this binomial distribution's $n$ and $p$. Jennifer is going to play five times, so $n$ is $5$. Every time she plays, she has a $25\%$ chance of winning.Following the conditional probability formula, let's let event $A$ be the event that she wins twice, and $B$ be the event that she wins at least once. That way, $\mathrm{Pr} (A | B)$ (in words, "the probability of $A$ given $B$") is the probability that she wins twice given that she wins at least once.$\mathrm{Pr} (B)$ is the probability that she wins at least once. This is a range question that can be answered with CDF expressions. There is also a complement shortcut since winning at least once covers every possible outcome except winning zero times.$$\mathrm{Pr}(B) = 1 - \mathrm{PDF}(0)$$Using a calculator, we see that $\mathrm{Pr}(B) \approx 0.7627$.Now, we need to calculate $\mathrm{Pr}(A \cap B)$. This is the probability that she wins twice AND wins more than once. Since winning twice already causes her to win more than once, this is a case where $\mathrm{Pr}(A \cap B)$ is actually the same as $\mathrm{Pr} (A)$.$$\mathrm{Pr}(A) = \mathrm{PDF}(2) \approx 0.2637$$Now we have all the pieces to calculate the answer to the question.$$\mathrm{Pr}(A | B) = \frac{\mathrm{Pr}(A \cap B)}{\mathrm{Pr}(B)}$$$$=\frac{0.2637}{0.7627}$$$$=0.3457$$In words, there is approximately a $37.6\%$ chance that Jennifer wins twice, when we consider only the outcomes where she won at all.

Put It To The Test

Give a few descriptive problems a try, and make sure to check out the practice worksheet for this lesson!
Example 5$X$ is a binomial random variable with $n=4$ and $p=0.1$. Determine the probability that $X$ is $0$.
Show solution
$\blacktriangleright$ This question wants to know about a probability associated with one outcome, not a range of outcomes, so we only need a PDF calculation.Pick up your calculator, find the binompdf option, and ask for the outcome with $n=4$, $p=0.1$, and $x=0$.If you are not allowed to use a graphing calculator, or if you have to show the work, we can get this answer with the formula.$$\mathrm{Pr}(X=0) = {4 \choose 0} (0.1)^{0} (0.9)^{4}$$Either way you should get $0.6561$, or about $66\%$.
Example 6The random variable $X$ represents the number of successes of a binomial experiment with $5$ trials and a success probability of $42\%$ for each trial. Calculate the following:a) The probability that $X$ is $3$b) $\mathrm{Pr}(X \lt 4)$c) The probability that there is at least $2$ successesd) The expected number of successes and the standard deviation of the number of successes
Show solution
$\blacktriangleright$ From the given info, we have a binomial outcome with $n=5$ trials, and trial success probability $p=0.42$. The computational challenge for each question will be in the setup.a) This is a PDF calculation. We want PDF($3$).$$\mathrm{Pr}(X=3) = {5 \choose 3} (0.42)^{3} (0.58)^{2}$$or, on a TI calculator, binompdf($n=5$,$p=0.42$,$x=3$).Answer: $\approx 0.2492$.b) This probability represents a range of outcomes, we we'll need a CDF. We want the probability that $X$ is less than $4$, so the CDF of $3$ will give us exactly that, remembering that CDF($3$) includes $3$.On the TI calculator, get binomcdf($n=5$,$p=0.42$,$x=3$).Answer: $\approx 0.8967$.Note that the only way to do this by hand is to add up individual PDFs. Most teachers don't go down this road anymore since it doesn't add value to your understanding but rather just takes time and is tedious to calculate.$$\mathrm{CDF}(3) = \mathrm{PDF}(0) + \mathrm{PDF}(1)$$$$+ \mathrm{PDF}(2) + \mathrm{PDF}(3)$$Or, slightly less work (but not substantially):$$\mathrm{CDF}(3) = 1 - \mathrm{PDF}(5)$$$$- \mathrm{PDF}(4)$$c) The probability that there are at least two successes is $\mathrm{Pr}(X \ge 2)$. One standard way you can approach "greater than" type questions is to always use the complement property of probability with CDF expressions, since CDF expressions will always represent "less than" type questions on their own.$$\mathrm{Pr}(X \ge 2) = 1 - \mathrm{Pr}(X \lt 2)$$$$=1 - \mathrm{Pr}(X \leq 1)$$$$=1 - \mathrm{CDF}(1)$$Now, go to your calculator to get the value of that CDF.binomcdf($n=5$,$p=0.42$,$x=1$) $\approx 0.3033$, so $1 - \mathrm{CDF}(1) \approx 0.6967$.d) For a binomial random variable $X$ with $n=5$ and $p=0.42$, we know that the expected value of $X$ is$$\mathrm{E}(X) = np = 5(0.42)$$$$=2.1$$and, the standard deviation is the square root of the variance, where the variance for a binomial variable is $npq$.$$\mathrm{SD}(X) = \sqrt{npq} = \sqrt{5(0.42)(0.58)}$$$$\approx 1.1036$$
Example 7A student did not study for a multiple choice exam and elects to randomly guess on each question. The exam has $15$ questions and each question has one correct answer choice and three incorrect answer choices. Find the expected number of questions he will answer correctly, and then find the probability that he answers enough questions to exceed the expected number.
Show solution
$\blacktriangleright$ Assuming this student is truly guessing randomly on each question, we can verify that this situation is binomial. After this important check is done, first jot down the $n$ and $p$ values.It is clear that $n$ is $15$, but we have to figure out $p$ for ourselves. Each question looks to have four total answer choices, one of which is correct, so $p$ is $1/4$ or $0.25$ (saying $p$ is $1/3$ is a common mistake for those who read too quickly - be careful!).The expected value of the number of questions he will answer correctly is then$$\mathrm{E}(X) = np = 15(0.25)$$$$=3.75$$Finally, we were also asked to calculate the probability that he exceeds the expected number. This is a very good chance to once again point out the difference between theoretical expectation and practical possibility! Recall from earlier discussion above and also from the prior lesson on discrete probability distributions » that expected values represent an average outcome if we were to do the whole $5$-draw experiment millions of times. It doesn't mean that we can necessarily get that exact expectation on any one experiment.Since this student is only able to answer an integer number of questions correctly, he will have to answer $4$ questions correctly in order to exceed the expected value of $3.75$. Set up an expression that represents this:$$\mathrm{Pr}(X \geq 4) = 1 - \mathrm{CDF}(3)$$The TI says that $\mathrm{CDF}(3)$ is approximately $0.4613$, so $\mathrm{Pr}(X \geq 4) \approx 0.5387$.
Example 8A representative sample of nine students is to be selected randomly from a large student body. If the probability of selecting a male or female is assumed to be exactly equal for each selection, what is the probability that the sample contains more males than females?
Show solution
$\blacktriangleright$ Similar to the last question, we must select our own $p$ and set up our own probability statement that aligns with what is being asked. If we believe that the student body is sufficiently numerous, we can believe that the probability of selecting a male is $0.5$.Because this is a sample of nine students, we will have more males than females if there are $5$ or more males.$$\mathrm{Pr}(X \geq 5) = 1 - \mathrm{CDF}(4)$$Calculator output tells us that $\mathrm{CDF}(4)$ is exactly $0.5$, so $1-0.5$ is also $0.5$, which is the answer.(see if you can think of why this result makes intuitive sense!)Note that for this problem we implicitly defined the outcome of each individual student selection to be a "success" if the choice was male, but we could have just as easily chosen female selection to be the "success". I encourage you to set up such problem where you can make a choice in whatever way best suits the question being asked. This question asked about probability of males exceeding females so we let our $X$ represent number of males selected, thus defining what we should let $p$ represent. Which choice you make would have mattered significantly more if the gender probabilities were not equal.
Lesson Takeaways
  • Understand binomial outcomes as groups of success / failure trials
  • Know what assumptions need to be validated for a success / failure situation to actually be binomial
  • Be able to set up word problems using PDF and CDF expressions
  • Know how to use your calculator to get whatever binomial probabilities you seek
  • Be well-practiced at the common types of questions you could be asked on a test, and the tricky word choices teachers like to use

Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

Key Lesson Sections

Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.

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