5.2: Binomial Probability Distribution

 

Learning Objectives

• Define the binomial probability distribution to model the number of successes in a fixed number of repeated, independent trials with two possible outcomes (success or failure).
• Calculate the probability of achieving a specific number of successes.
• Use the total number of trials, the probability of success, and the desired number of successes in the computation.

Section 5.1 introduced the concept of a probability distribution. The focus of the section was on discrete probability distributions (PDF). The PDF is found by conducting an experiment to collect data and then computing the experimental probabilities. Normally, the theoretical probabilities cannot be computed. However, certain types of experiments allow for the calculation of the theoretical probability. One of those types is called a Binomial Experiment.

Properties of a binomial experiment (or Bernoulli trial)

1. Fixed number of trials, n, which means that the experiment is repeated a specific number of times.
2. The n trials are independent, which means that what happens on one trial does not influence the outcomes of other trials.
3. There are only two outcomes, which are called a success and a failure.
4. The probability of success doesn’t change from trial to trial, where p = probability of success and q = probability of failure, q = 1-p.

Once a binomial experiment has been verified, then binomial probabilities can be calculated. This is important because binomial probabilities come up often in real life. Examples of binomial experiments are:

• Toss a fair coin ten times, and find the probability of getting two heads.
• Question twenty people in class, and look for the probability of more than half being women.
• Shoot five arrows at a target, and find the probability of hitting it five times.

To develop the process for calculating the probabilities in a binomial experiment, consider Example \(\PageIndex{1}\).

Binomial Distribution Formula

The random variable in a binomial experiment is x = number of successes. However, success is not always a good thing. Sometimes, success is bad, like finding a defect. Success means you observed the outcome you wanted to see happen.

Definition \(\PageIndex{1}\)

The binomial Formula for the probability of x successes in n trials is

\(P(x)=_{n} C_{x} p^{x} q^{n \cdot x} \text { where }_{n} C_{x}=\dfrac{n !}{x !(n-x) !}\)

The \(_{n} C_{x}\) is the number of combinations of n things taking x at a time. It is read “n choose x”.

When solving problems, make sure you define your random variable and state what n, p, q, and x are. Without doing this, the problems are a great deal harder.

Example \(\PageIndex{1}\): Calculating Binomial Probabilities

When looking at a person’s eye color, it turns out that 1% of people in the world have green eyes ("What percentage of," 2013). Consider a group of 20 people. Round the results to three decimal places.

1. State the random variable.
2. Verify that this is a binomial experiment.
3. Find the probability that none have green eyes.
4. Find the probability that exactly nine have green eyes.
5. Find the probability that at most one has green eyes.
6. Find the probability that at least two have green eyes.

Solution

1. x = number of people with green eyes
2. Verification is presented below.
   • There are 20 people, and each person is a trial, so there is a fixed number of trials. In this case, \(n = 20\).
   • If you assume that each person in the group is chosen at random, the eye color of one person doesn’t affect the eye color of the next person, thus, the trials are independent.
   • Either a person has green eyes or they do not have green eyes, so there are only two outcomes. In this case, success is when a person has green eyes.
   • The probability of a person having green eyes is 0.01. This is the same for every trial since each person has the same chance of having green eyes. p = 0.01 and q = 1 - 0.01 = 0.99
3. \(P(x=0)=_{20} C_{0}(0.01)^{0}(0.99)^{20-0} \approx 0.818\)
4. \(P(x=9)=_{20} C_{9}(0.01)^{9}(0.99)^{20-9} \approx 1.50 \times 10^{-13} \approx 0.000\)
5. At most one means that one is the highest value you will have. Find the probability that x is less than or equal to three.

\(\begin{aligned} P(x \leq 1) &=P(x=0)+P(x=1) \\ &=_{20} C_{0}(0.01)^{0}(0.99)^{20}+_{20} C_{1}(0.01)^{1}(0.99)^{19} \\& = 0.818+0.165 = 0.983 \end{aligned}\)

6. To find the probability that at least two people have green eyes, it is simpler to calculate it as \( 1 - P(\text{At Most 1})\).

\(\begin{aligned} P(x \geq 2) &=1 - P(x \leq 1) \\ &=1 - 0.983 \\ & =0.017 \end{aligned}\)

The binomial formula is cumbersome to use, and probabilities can be easily computed using technology. On the TI-83/84 calculator, the commands on the TI-83/84 calculators when the number of trials is equal to n and the probability of a success is equal to p are \(\text{binompdf}(n, p, r)\) when you want to find P(x=r) and \(\text{binomcdf}(n, p, r)\) when you want to find \(P(x \leq r)\). If you want to find \(P(x \geq r)\), then you use the property that \(P(x \geq r)=1-P(x \leq r-1)\), since \(x \geq r\) and \(x<r\) or \(x \leq r-1\) are complementary events. Both binompdf and binomcdf commands are found in the [DISTR] menu.

Example \(\PageIndex{2}\) Using the Binomial Probability Table

A fair coin is flipped 10 times. Count how many times heads appear. Since each flip is independent with two outcomes, this is a binomial situation with a probability of 0.5 for heads. Use a binomial probability table to find the following probabilities.

1. Get exactly 6 heads.
2. Get at least 6 heads.
3. Get at most 3 heads.

Solution

Identify n and p; these values are n = 10 (number of trials) and p = 0.5 (probability of success).

Go to chapter 13.1: The Binomial Distribution Table for n = 2 to n = 10. In the table, search for n = 10 and p = 0.5. The column where these intersect will contain the probabilities.

1. Go to n = 10, X = 6, then find the probability in the column where p = 0.5. From the table, the probability of exactly 6 successes is P(6) = 0.205.
2. At least 6 translates to 6 or more. Go to n = 10, find all the probabilities from X = 6 until 10 under p =0.5. Add these probabilities together. Also, since we are finding the probability of at least 6, then we use the notation P( X \( \ge \) 6). Thus, the answer is P( X \( \ge \) 6) = P(6) + P(7) + P(8) + P(9) + P(10) = 0.205 + 0.117 + 0.044 + 0.010 + 0.001 = 0.377.
3. At most 3 translates to 3 or fewer. Go to n = 10, find all the probabilities from X = 0 until 3 under p =0.5. Add these probabilities together. Also, since we are finding the probability of at most 3, then we use the notation P( X \( \le \) 3). Thus, the answer is P( X \( \le \) 3) = P(0) + P(1) + P(2) + P(3) = 0.001 + 0.010 + 0.044 + 0.017 = 0.172.

Example \(\PageIndex{3}\) Using the Binomial Command on the TI-83/84

When looking at a person’s eye color, it turns out that 1% of people in the world have green eyes ("What percentage of," 2013). Consider a group of 20 people.

1. State the random variable.
2. Find the probability that none have green eyes.
3. Find the probability that exactly nine have green eyes.
4. Find the probability that at most three have green eyes.
5. Find the probability that at most two have green eyes.

Solution

a. x = number of people with green eyes

b. You are looking for P (x=0). Since this problem is x=0, you use the binompdf command on the TI-83/84. On the TI83/84, you go to the [DISTR] menu, select the binompdf, and then type into the parentheses your n, p, and x values into your calculator, making sure you use the comma to separate the values. The command will look like \(\text{binompdf}(20,.01,0)\) and when you press [ENTER] you will be answered. (If you have the new software on the TI-84, the screen looks a bit different.)

Figure \(\PageIndex{1}\): Calculator Results for binompdf

P (x=0) = 0.8179. Thus, there is an 81.8% chance that in a group of 20 people, none of them will have green eyes.

c. In this case, you want to find the P (x=9). Again, you will use the binompdf command or the dbinom command. Following the procedure above, you will have binompdf(20, .01, 9) on the TI-83/84. Your answer is \(P(x=9)=1.50 \times 10^{-13}\). (Remember when the calculator gives you \(1.50 E-13\), this is how they display scientific notation.) The probability that out of twenty people, nine have green eyes is a very small chance.

d. At most three means that three is the highest value you will have. Find the probability of x being less than or equal to three, which is \(P(x \leq 3)\). This uses the binomcdf command on the TI-83/84. You use the command on the TI-83/84 of binomcdf(20, .01, 3).

Figure \(\PageIndex{2}\): Calculator Results for binomcdf

Your answer is 0.99996. Thus, there is a good chance that in a group of 20 people, at most three will have green eyes. (Note: don’t round this to one, since one means that the event will happen when there is a slight chance that it won’t happen. It is best to write the answer out to enough decimal points so it doesn’t round off to one.

e. You are looking for \(P(x \leq 2)\). Again use binomcdf or pbinom. Following the procedure above you will have \(\text{binomcdf}(20,.01,2)\) on the TI-83/84 and pbinom(2,20,0.01), with \(P(x \leq 2)=0.998996\). Again, there is a good chance that at most two people in the room will have green eyes.

Example \(\PageIndex{4}\) Calculating Binomial Probabilities

According to the Centers for Disease Control (CDC), about 1 in 88 children in the U.S. have been diagnosed with autism ("CDC-data and statistics," 2013). Suppose you consider a group of 10 children.

1. State the random variable.
2. Verify that this is a binomial experiment.
3. Find the probability that none have autism.
4. Find the probability that exactly seven have autism.
5. Find the probability that at least five have autism.
6. Find the probability that at most two have autism.
7. Suppose five children out of ten have autism. Is this unusual? What does that tell you?

Solution

a. x = number of children with autism

b.

1. There are 10 children, and each child is a trial, so there is a fixed number of trials. In this case, n = 10.
2. If you assume that each child in the group is chosen at random, then whether a child has autism does not affect the chance that the next child has autism. Thus, the trials are independent.
3. Either a child has autism or they do not have autism, so there are two outcomes. In this case, the success is that a child has autism.
4. The probability of a child having autism is 1/88. This is the same for every trial since each child has the same chance of having autism. \(p=\dfrac{1}{88}\) and \(q=1-\dfrac{1}{88}=\dfrac{87}{88}\).

c. Using the formula:

\(P(x=0)=_{10} C_{0}\left(\dfrac{1}{88}\right)^{0}\left(\dfrac{87}{88}\right)^{10-0} \approx 0.892\)

Using the TI-83/84 Calculator:

\(P(x=0)=\text { binompdf }(10,1 \div 88,0) \approx 0.892\)

d. Using the formula:

\(P(x=7)=_{10} C_{7}\left(\dfrac{1}{88}\right)^{7}\left(\dfrac{87}{88}\right)^{10-7} \approx 0.000\)

Using the TI-83/84 Calculator:

\(P(x=7)=\text { binompdf }(10,1 \div 88,7) \approx 2.84 \times 10^{-12}\)

e. Using the formula:

\(\begin{aligned} P(x \geq 5) &=P(x=5)+P(x=6)+P(x=7) \\ &+P(x=8)+P(x=9)+P(x=10) \\ &=_{10} C_{5}\left(\dfrac{1}{88}\right)^{5}\left(\dfrac{78}{88}\right)^{10-5}+_{10} C_{6}\left(\dfrac{1}{88}\right)^{6}\left(\dfrac{78}{88}\right)^{10-6} \\ & +_{10}C_{7}\left(\dfrac{1}{88}\right)^{7}\left(\dfrac{78}{88}\right)^{10-7}+_{10}C_{8}\left(\dfrac{1}{88}\right)^{8}\left(\dfrac{78}{88}\right)^{10-8} \\ &+_{10}C_{9}\left(\dfrac{1}{88}\right)^{9}\left(\dfrac{78}{88}\right)^{10-9}+_{10}C_{10}\left(\dfrac{1}{88}\right)^{10}\left(\dfrac{78}{88}\right)^{10-10}\\&=0.000+0.000+0.000+0.000+0.000+0.000 \\ &=0.000 \end{aligned}\)

Using the TI-83/84 Calculator:

To use the calculator, you need to use the complement.

\(\begin{aligned} P(x \geq 5) &=1-P(x<5) \\ &=1-P(x \leq 4) \\ &=1-\text { binomcdf }(10,1 \div 88,4) \\ & \approx 1-0.9999999=0.000 \end{aligned}\)

Notice, the answer is given as 0.000, since the answer is less than 0.000. Don’t write 0, since 0 means that the event is impossible to happen. The event of five or more is improbable, but not impossible.

f. Using the formula:

\(\begin{aligned} P(x \leq 2) &=P(x=0)+P(x=1)+P(x=2) \\ &=_{10} C_{0}\left(\dfrac{1}{88}\right)^{0}\left(\dfrac{78}{88}\right)^{10-0}+_{10} C_{1}\left(\dfrac{1}{88}\right)^{1}\left(\dfrac{78}{88}\right)^{10-1} \\ &+_{10} C_{2}\left(\dfrac{1}{88}\right)^{2}\left(\dfrac{78}{88}\right)^{10-2} \\ &=0.892+0.103+0.005>0.999 \end{aligned}\)

Using the TI-83/84 Calculator:

\(P(x \leq 2)=\text { binomcdf }(10,1 \div 88,2) \approx 0.9998\)

g. Since the probability of five or more children in a group of ten having autism is much less than 5%, it is unusual to happen. If this does happen, then one may think that the proportion of children diagnosed with autism is more than 1/88.

Example \(\PageIndex{5}\) Calculating Binomial Probabilities

A university study finds that 60% of college students regularly drink coffee to stay focused while studying. Suppose we randomly select 10 college students.

1. What is the probability that exactly 7 of the students drink coffee regularly?
2. What is the probability that fewer than 3 students drink coffee regularly?
3. What is the probability that more than 8 students drink coffee regularly?

Solution

1. For this problem \( n = 10, p = 0.6, and q = 1 - 0.6 = 0.4\).

\(\begin{aligned} P(X = 7) &= _{10} C_{7} \cdot (0.6)^{7} (0.4)^{3} \\ &= 0.215 \end{aligned}\)

2. For the problem, all the probabilities P(0), P(1), and P(2) must be computed and added together.

\(\begin{aligned} P(X < 3) &= P(0) + P(1) +P(2) \\ & = _{10} C_{0} \cdot (0.6)^{0} (0.4)^{10} + _{10} C_{1} \cdot (0.6)^{1} (0.4)^{9} + _{10} C_{2} \cdot (0.6)^{2} (0.4)^{8}\\ &= 0.000+ 0.002 + 0.011 \\ & = 0.013\end{aligned}\)

3. For the problem, all the probabilities P(9) and P(10) must be computed and added together.

\(\begin{aligned} P(X > 8) &= P(X = 9) + P(X = 10) \\ &= _{10} C_{9} \cdot (0.6)^{9} (0.4)^{1} + _{10} C_{10} \cdot (0.6)^{10} (0.4)^{0} \\ &= 0.040 + 0.006 \\ &= 0.046 \end{aligned}\)

Authors

"5.2: Binomial Probability Distribution" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0

Attributions

"5.2: Binomial Probability Distribution" by Kathryn Kozak is licensed under CC BY-SA 4.0