Section3.3The binomial formula

What is the probability of exactly 50 heads in 100 coin tosses? Or the probability of randomly sampling 8 people and having exactly 1 of them be left-handed? Or of at most 3 people exceeding their insurance deductible in a random sample of 20 people? The binomial formula can help us answer these questions.

Subsection3.3.1Learning objectives

1. Calculate the number of possible scenarios for obtaining $x$ successes in $n$ trials.

2. Determine whether a scenario is binomial or not.

3. Calculate the probability of obtaining exactly $x$ successes in $n$ independent trials.

4. Recognize that the binomial formula uses the special Addition Rule for mutually exclusive events.

5. Find probabilities of the form “at least” or “at most” by applying the binomial formula multiple times.

Subsection3.3.2Introducing the binomial formula

Many health insurance plans in the United States have a deductible, where the insured individual is responsible for costs up to the deductible, and then the costs above the deductible are shared between the individual and insurance company for the remainder of the year.

Suppose a health insurance company found that 70% of the people they insure stay below their deductible in any given year. Each of these people can be thought of as a trial. We label a person a success if her healthcare costs do not exceed the deductible. We label a person a failure if she does exceed her deductible in the year. Because 70% of the individuals will not hit their deductible, we denote the probability of a success as $p = 0.7\text{.}$

Suppose the insurance agency is considering a random sample of four individuals they insure. What is the chance exactly one of them will exceed the deductible and the other three will not? Let's call the four people Ariana ($A$), Brittany ($B$), Carlton ($C$), and Damian ($D$) for convenience.

Solution

Let's consider a scenario where one person exceeds the deductible:

\begin{align*} \amp P(A=\text{exceed, } B= \text{not, } C=\text{not, } D= \text{not} )\\ \amp = P(A=\text{exceed} )\ P(B=\text{not} )\ P(C=\text{not} )\ P(D=\text{not} )\\ \amp = (0.3)(0.7)(0.7) (0.7)\\ \amp = (0.7)^3 (0.3)^1\\ \amp = 0.103 \end{align*}

But there are three other scenarios: Brittany, Carlton, or Damian could have been the one to exceed the deductible. In each of these cases, the probability is again $(0.7)^3 (0.3)^1\text{.}$ These four scenarios exhaust all the possible ways that exactly one of these four people could have exceeded the deductible, so the total probability is $4 \times (0.7)^3 (0.3)^1 =0.412\text{.}$

Verify that the scenario where Brittany is the only one to exceed the deductible has probability $(0.7)^3 (0.3)^1\text{.}$  1

$P(A= \text{not, } B= \text{exceed, } C=\text{not, } D=\text{not}) =(0.7)(0.3)(0.7)(0.7)=(0.7)^3(0.3)^1$

The binomial distribution describes the probability of having exactly $x$ successes in $n$ independent trials with probability of a success $p$ (in Example 3.3.1.1, $n=4\text{,}$ $x=3\text{,}$ $p=0.7$). We would like to determine the probabilities associated with the binomial distribution more generally, i.e. we want a formula where we can use $n\text{,}$ $x\text{,}$ and $p$ to obtain the probability. To do this, we reexamine each part of Example 3.3.1.1.

There were four individuals who could have been the one to exceed the deductible, and each of these four scenarios had the same probability. Thus, we could identify the final probability as

\begin{gather*} [ \text{# of scenarios}] \times P(\text{single scenario} ) \end{gather*}

The first component of this equation is the number of ways to arrange the $x=3$ successes among the $n=4$ trials. The second component is the probability of any of the four (equally probable) scenarios.

Consider $P(\text{single scenario})$ under the general case of $x$ successes and $n-x$ failures in the $n$ trials. In any such scenario, we apply the Multiplication Rule for independent events:

\begin{gather*} p^x (1 - p)^{n - x} \end{gather*}

This is our general formula for $P(\text{single scenario})\text{.}$

Secondly, we introduce the binomial coefficient, which gives the number of ways to choose $x$ successes in $n$ trials, i.e. arrange $x$ successes and $n - x$ failures:

\begin{gather*} {n \choose x} = \frac{n!}{x! (n - x)!} \end{gather*}

The quantity ${n \choose x}$ is read n choose x. 2  The exclamation point notation (e.g. $n!$) denotes a factorial expression.

\begin{align*} \amp 0! = 1\\ \amp 1! = 1\\ \amp 2! = 2\times1 = 2\\ \amp 3! = 3\times2\times1 = 6\\ \amp 4! = 4\times3\times2\times1 = 24\\ \amp \vdots\\ \amp n! = n\times(n-1)\times...\times3\times2\times1 \end{align*}
Other notations for $n$ choose $x$ includes $_nC_x\text{,}$ $C_n^x\text{,}$ and $C(n,x)\text{.}$

Using the formula, we can compute the number of ways to choose $x = 3$ successes in $n = 4$ trials:

\begin{gather*} {4 \choose 3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4\times3\times2\times1}{(3\times2\times1) (1)} = 4 \end{gather*}

This result is exactly what we found by carefully thinking of each possible scenario in Example 3.3.1.1.

Substituting $n$ choose $x$ for the number of scenarios and $p^x(1-p)^{n-x}$ for the single scenario probability yields the binomial formula.

Binomial formula.

Suppose the probability of a single trial being a success is $p\text{.}$ Then the probability of observing exactly $x$ successes in $n$ independent trials is given by

\begin{gather*} P(X=x)={n\choose x}p^x(1-p)^{n-x} = \frac{n!}{x!(n-x)!}p^x(1-p)^{n-x} \end{gather*}

Subsection3.3.3When and how to apply the formula

Is it binomial? Four conditions to check..

(1) The trials are independent.

(2) The number of trials, $n\text{,}$ is fixed.

(3) Each trial outcome can be classified as a success or failure.

(4) The probability of a success, $p\text{,}$ is the same for each trial.

What is the probability that 3 of 8 randomly selected individuals will have exceeded the insurance deductible, i.e. that 5 of 8 will not exceed the deductible? Recall that 70% of individuals will not exceed the deductible.

Solution

We would like to apply the binomial model, so we check the conditions. The number of trials is fixed ($n = 8$) (condition 2) and each trial outcome can be classified as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4).

In the outcome of interest, there are $x = 5$ successes in $n = 8$ trials (recall that a success is an individual who does not exceed the deductible, and the probability of a success is $p = 0.7\text{.}$ So the probability that 5 of 8 will not exceed the deductible and 3 will exceed the deductible is given by

\begin{align*} { 8 \choose 5}(0.7)^5 (1-0.7)^{8-5} \amp = \frac{8!}{5!(5-3)!} (0.7)^5(1-0.7)^{8-5}\\ \amp = \frac{8!}{5!3!} (0.7)^5(0.3)^3 \end{align*}

Dealing with the factorial part:

\begin{gather*} \frac{8!}{5!3!} = \frac{8\times7\times6\times5\times4\times3\times2\times1} {(5\times4\times3\times2\times1)(3\times2\times1)} = \frac{8\times7\times6}{3\times2\times1} = 56 \end{gather*}

Using $(0.7)^5(0.3)^3 \approx 0.00454\text{,}$ the final probability is about $56\times 0.00454 \approx 0.254\text{.}$

If you must calculate the binomial coefficient by hand, it's often useful to cancel out as many terms as possible in the top and bottom. See Section 3.3.4 for how to evaluate the binomial coefficient and the binomial formula using a calculator.

Computing binomial probabilities.

The first step in using the binomial model is to check that the model is appropriate. The second step is to identify $n\text{,}$ $p\text{,}$ and $x\text{.}$ Finally, apply the binomial formula to determine the probability and interpret the results.

Approximately 35% of a population has blood type O+. Suppose four people show up at a hospital and we want to find the probability that exactly one of them has blood type O+. Can we use the binomial formula?

Solution

To check if the binomial model is appropriate, we must verify the conditions.

1. If we suppose that these 4 people comprise a random sample, then we can treat them as independent. This seems reasonable, since one person with a particular blood type showing up at a hospital seems unlikely to affect the chance that other people with that blood type would show up at the hospital.

2. We have a fixed number of trials ($n=4$).

3. Each outcome is a success or failure (blood type O+ or not blood type O+).

4. The probability of a success is the same for each trial since the individuals are like a random sample ($p=0.35$ if we say a “success” is someone having blood type O+).

The probability that a random smoker will develop a severe lung condition in his or her lifetime is about $0.3\text{.}$ If you have 4 friends who smoke and you want to find the probability that 1 of them will develop a severe lung condition in his or her lifetime, can you apply the binomial formula?  3

While conditions (2) and (3) are met, most likely the friends know each other, so the independence assumption (1) is probably not satisfied. For example, acquaintances may have similar smoking habits, or those friends might make a pact to quit together. Condition (4) is also not satisfied since this is not a random sample of people.

Given that 35% of a population has blood type O+, what is the probabilty that in a random sample of 4 people:

1. none of them have blood type O+?

2. one will have blood type O+?

3. no more than one will have blood type O+?

Solution

Compute parts (a) and (b) using the binomial formula:

1. $P(X=0) = {4 \choose 0} (0.35)^0 (0.65)^4 = 1\times1\times 0.65^4 = 0.65^4 = 0.179$ Note that we could have answered this question without the binomial formula, using methods from the previous section.

2. $P(X=1) = {4 \choose 1} (0.35)^1(0.65)^{3} = 0.384\text{.}$

3. This can be computed as the sum of parts (a) and (b): $P(X=0) + P(X=1) = 0.179 + 0.384 = 0.563\text{.}$ That is, there is about a 56.3% chance that no more than one of them will have blood type O+.

What is the probability that at least 3 of 4 people in a random sample will have blood type O+ if 35% of the population has blood type O+?  4

$P(\text{at least 3 of 4 have blood type O+})=P(X=3)+P(X=4)={ 4 \choose 3}(0.35)^3(0.65)^1+(0.35)^4=0.111+0.015=0.126$

There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles without replacement. Find the probability you get exactly 3 blue marbles.

Solution

Because the probability of success $p$ is not the same for each trial, we cannot use the binomial formula. However, we can use the same logic to arrive at the following answer.

\begin{align*} P(X = 3) \amp = (\text{# of combinations with 3 blue } )\times P(\text{3 blue and 2 red in a specific order})\\ \amp ={5\choose 3}\times P(\text{ BBBRR } )\\ \amp = {5\choose 3}\left(\frac{4}{13}\times \frac{3}{12}\times \frac{2}{11} \times \frac{9}{10} \times \frac{8}{9}\right)\\ \amp = 0.1119 \end{align*}

Draw 4 cards without replacement from a deck of 52 cards. What is the probability that you get at least two hearts?  5

$P(\text{at least 2 hearts in 4 draws from a deck})=1-[P(X=0)+P(X=1)]=1-[(\frac{39}{52}) (\frac{38}{51}) (\frac{37}{50}) (\frac{36}{49})+{4\choose 1}(\frac{13}{52}) (\frac{39}{51}) (\frac{38}{50}) (\frac{37}{49})]=1-[0.03038+0.4388]=0.2574$

Lastly, we consider the binomial coefficient,, $n$ choose $x\text{,}$ under some special scenarios.

Why is it true that ${n \choose 0}=1$ and ${n \choose n}=1$ for any number $n\text{?}$  6

Frame these expressions into words. How many different ways are there to arrange 0 successes and $n$ failures in $n$ trials? (1 way.) How many different ways are there to arrange $n$ successes and 0 failures in $n$ trials? (1 way.)

How many ways can you arrange one success and $n-1$ failures in $n$ trials? How many ways can you arrange $n-1$ successes and one failure in $n$ trials?  7

One success and $n-1$ failures: there are exactly n unique places we can put the success, so there are $n$ ways to arrange one success and $n-1$ failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations:${n \choose 1}=n\text{,}$ ${n \choose {n-1}}=n$

Subsection3.3.4Calculator: binomial probabilities

TI-83/84: Computing the binomial coefficient ${n\choose x}$.

Use MATH, PRB, nCr to evaluate $n$ choose $r\text{.}$ Here $r$ and $x$ are different letters for the same quantity.

1. Type the value of $n\text{.}$

2. Select MATH.

3. Right arrow to PRB.

4. Choose 3:nCr.

5. Type the value of $x\text{.}$

6. Hit ENTER.

Example: 5 nCr 3 means 5 choose 3.

Casio fx-9750GII: Computing the binomial coefficient ${n\choose x}$.
1. Navigate to the RUN-MAT section (hit MENU, then hit 1).

2. Enter a value for $n\text{.}$

3. Go to CATALOG (hit buttons SHIFT and then 7).

4. Type C (hit the ln button), then navigate down to the bolded C and hit EXE.

5. Enter the value of $x\text{.}$ Example of what it should look like: 7C3.

6. Hit EXE.

TI-84: Computing the binomial formula, $P(X = {x)={n\choose x}p^x(1-p)^{n-x}}$.

Use 2ND VARS, binompdf to evaluate the probability of exactly $x$ occurrences out of $n$ independent trials of an event with probability $p\text{.}$

1. Select 2ND VARS (i.e. DISTR)

2. Choose A:binompdf (use the down arrow to scroll down).

3. Let trials be $n\text{.}$

4. Let p be $p$

5. Let x value be $x\text{.}$

6. Select Paste and hit ENTER.

TI-83: Do step 1, choose 0:binompdf, then enter $n\text{,}$ $p\text{,}$ and $x$ separated by commas:

binompdf(n, p, x). Then hit ENTER.

TI-84: Computing $P(X \le {x)= {n\choose 0}p^0(1-p)^{n-0} + ... + {n\choose x}p^x(1-p)^{n-x}}$.

Use 2ND VARS, binomcdf to evaluate the cumulative probability of at most $x$ occurrences out of $n$ independent trials of an event with probability $p\text{.}$

1. Select 2ND VARS (i.e. DISTR)

2. Choose B:binomcdf (use the down arrow).

3. Let trials be $n\text{.}$

4. Let p be $p$

5. Let x value be $x\text{.}$

6. Select Paste and hit ENTER.

TI-83: Do steps 1-2, then enter the values for $n\text{,}$ $p\text{,}$ and $x$ separated by commas as follows: binomcdf(n, p, x). Then hit ENTER.

Casio fx-9750GII: Binomial calculations.
1. Navigate to STAT (MENU, then hit 2).

2. Select DIST (F5), and then BINM (F5).

3. Choose whether to calculate the binomial distribution for a specific number of successes, $P(X = k)\text{,}$ or for a range $P(X \leq k)$ of values (0 successes, 1 success, ..., $x$ successes).

• For a specific number of successes, choose Bpd (F1).

• To consider the range 0, 1, ..., $x$ successes, choose Bcd(F1).

4. If needed, set Data to Variable (Var option, which is F2).

5. Enter the value for x ($x$), Numtrial ($n$), and p (probability of a success).

6. Hit EXE.

Find the number of ways of arranging 3 blue marbles and 2 red marbles.  8

Here c$=5$ and x$=3\text{.}$ Doing 5 nCr 3 gives the number of combinations as 10

There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get exactly 3 blue marbles.  9

Here, $n=5\text{,}$ $p=4/13$ and $x=3\text{,}$ so set trials$=5\text{,}$ p$=4/13$ and x value$=3\text{.}$ The probability is 0.1396.

There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get at most 3 blue marbles (i.e. less than or equal to 3 blue marbles).  10

Similarly, set trials$=5\text{,}$ p$=4/13$ and x value$=3\text{.}$ The cumulative probability is 0.9662

Subsection3.3.5Section summary

• $n\choose x\text{,}$ the binomial coefficient, describes the number of combinations for arranging $x$ successes among $n$ trials. $n\choose x$ $=\frac{n!}{x!(n-x)!}\text{,}$ where $n!=1\times 2\times 3\times...n\text{,}$ and $0!=0\text{.}$

• The binomial formula can be used to find the probability that something happens exactly x times in n trials. Suppose the probability of a single trial being a success is $p\text{.}$ Then the probability of observing exactly $x$ successes in $n$ independent trials is given by

\begin{gather*} {n\choose x}p^x(1-p)^{n-x} = \frac{n!}{x!(n-x)!}p^x(1-p)^{n-x} \end{gather*}
• To apply the binomial formula, the events must be independent from trial to trial. Additionally, $n\text{,}$ the number of trials must be fixed in advance, and $p\text{,}$ the probability of the event occurring in a given trial, must be the same for each trial.

• To use the binomial formula, first confirm that the binomial conditions are met. Next, identify the number of trials $n\text{,}$ the number of times the event is to be a “success” $x\text{,}$ and the probability that a single trial is a success $p\text{.}$ Finally, plug these three numbers into the formula to get the probability of exactly $x$ successes in $n$ trials.

• The $p^x(1-p)^{n-x}$ part of the binomial formula is the probability of just one combination. Since there are $n\choose x$ combinations, we add $p^x(1-p)^{n-x}$ up $n\choose x$ times. We can think of the binomial formula as: $[\# \text{ of combinations } ] \times P(\text{ a single combination } )\text{.}$

• To find a probability involving at least or at most, first determine if the scenario is binomial. If so, apply the binomial formula as many times as needed and add up the results. e.g. $P(\text{ at least 3 Heads in 5 tosses of a fair coin } )=P(\text{ exactly 3 Heads } )+P(\text{ exactly 4 Heads } )+P(\text{ exactly 5 Heads } )\text{,}$ where each probability can be found using the binomial formula.

Exercises3.3.6Exercises

1.Exploring combinations.

A coin is tossed 5 times. How many sequences / combinations of Heads/Tails are there that have:

1. Exactly 1 Tail?

2. Exactly 4 Tails?

3. Eactly 3 Tails?

4. At least 3 Tails?

Solution

(a) ${5 \choose 1}=5\text{.}$

(b) ${5 \choose 4}=5\text{.}$

(c) ${5 \choose 3}=10\text{.}$

(d) ${5 \choose 3} +{5 \choose 4}+ {5 \choose 5}=10+5+1 =16\text{.}$

2.Political affiliation.

Suppose that in a large population, 51% identify as Democrat. A researcher takes a random sample of 3 people.

1. Use the binomial model to calculate the probability that two of them identify as Democrat.

2. Write out all possible orderings of 3 people, 2 of whom identify as Democrat. Use these scenarios to calculate the same probability from part (a) but using the Addition Rule for disjoint events. Confirm that your answers from parts (a) and (b) match.

3. If we wanted to calculate the probability that a random sample of 8 people will have 3 that identify as Democrat, briefly describe why the approach from part (b) would be more tedious than the approach from part (a).

3.Underage drinking, Part I.

Data collected by the Substance Abuse and Mental Health Services Administration (SAMSHA) suggests that 69.7% of 18-20 year olds consumed alcoholic beverages in any given year. 11

1. Suppose a random sample of ten 18-20 year olds is taken. Is the use of the binomial distribution appropriate for calculating the probability that exactly six consumed alcoholic beverages? Explain.

2. Calculate the probability that exactly 6 out of 10 randomly sampled 18- 20 year olds consumed an alcoholic drink.

3. What is the probability that exactly four out of ten 18-20 year olds have not consumed an alcoholic beverage?

4. What is the probability that at most 2 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages?

5. What is the probability that at least 1 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages?

SAMHSA, Office of Applied Studies, National Survey on Drug Use and Health, 2007 and 2008.
Solution

(a) Yes. The conditions are satisfied: independence, fixed number of trials, either success or failure for each trial, and probability of success being constant across trials.

(b) 0.200.

(c) 0.200.

(d) $0.0024 + 0.0284 + 0.1323 = 0.1631\text{.}$

(e) $1 - 0.0024 = 0.9976\text{.}$

4.Chicken pox, Part I.

The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood. 12

1. Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood? Explain.

2. Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.

3. What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood?

4. What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?

5. What is the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox?

National Vaccine Information Center, Chickenpox, The Disease & The Vaccine Fact Sheet