Binomial Probability

In this section, we will consider types of problems that involve a sequence of trials, where each trial has only two outcomes, a success or a failure. These trials are independent. That is, the outcome of one does not affect the outcome of any other trial. Furthermore, the probability of success, p, and the probability of failure, (1 − p), remains the same throughout the experiment. These problems are called binomial probability problems. Since these problems were researched by a Swiss mathematician named Jacques Bernoulli around 1700, they are also referred to as Bernoulli trials.

We give the following definition:

Binomial Experiment: A binomial experiment satisfies the following four conditions:

  1. There are only two outcomes, a success or a failure, for each trial.
  2. The same experiment is repeated several times.
  3. The trials are independent; that is, the outcome of a particular trial does not affect the outcome of any other trial.
  4. The probability of success remains the same for every trial.

 

The probability model that we are about to investigate will give us the tools to solve many real-life problems like the ones given below.

  1. If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?
  2. If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 7 out of 10 free throws in a game?
  3. If a medicine cures 80% of the people who take it, what is the probability that among the 10 people who take the medicine, 6 will be cured?
  4. If a microchip manufacturer claims that only 4% of their chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective?
  5. If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product?

We now consider the following example to develop a formula for finding the probability of k successes in n Bernoulli trials.

 

Example 7.1.1

A baseball player has a batting average of 0.300. If he bats four times in a game, find the probability that he will have:
a. four hits
b. three hits
c. two hits
d. one hit
e. no hits.
Solution
Let us suppose S denotes that the player gets a hit, and F denotes that he does not get a hit. This is a binomial experiment because it meets all four conditions. First, there are only two outcomes, S or F. Clearly the experiment is repeated four times. Lastly, if we assume that the player’s skillfulness to get a hit does not change each time he comes to bat, the trials are independent with a probability of 0.3 of getting a hit during each trial. We draw a tree diagram to show all situations.

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Let us first find the probability of getting, for example, two hits. We will have to consider the six possibilities, SSFF , SFSF , SFFS , FSSF , FSFS , FFSS , as shown in the above tree diagram. We list the probabilities of each below.
P(SSFF) = (0.3)(0.3)(0.7)(0.7) = (0.3)2(0.7)2
P(SFSF) = (0.3)(0.7)(0.3)(0.7) = (0.3)2(0.7)2
P(SFFS) = (0.3)(0.7)(0.7)(0.3) = (0.3)2(0.7)2
P(FSSF) = (0.7)(0.3)(0.3)(0.7) = (0.3)2(0.7)2
P(FSFS) = (0.7)(0.3)(0.7)(0.3) = (0.3)2(0.7)2
P(FFSS) = (0.7)(0.7)(0.3)(0.3) = (0.3)2(0.7)2
Since the probability of each of these six outcomes is (0.3)2(0.7)2, the probability of obtaining two successes is 6(0.3)2(0.7)2.
The probability of getting one hit can be obtained in the same way. Since each permutation has one S and three F‘s, there are four such outcomes: SFFF , FSFF , FFSF , and FFFS.
And since the probability of each of the four outcomes is (0.3)(0.7)3, the probability of getting one hit is 4(0.3)(0.7)3.
The table below lists the probabilities for all cases, and shows a comparison with the binomial expansion of fourth degree. Again, p denotes the probability of success, and q = (1 − p) the probability of failure.
Outcome Four Hits Three hits Two Hits One hits No Hits
Probability (0.3)4 4(0.3)3(0.7) 6(0.3)2(0.7)2 4(0.3)(0.7)3 (0.7)4

 

This gives us the following theorem:

Binomial Probability Theorem:

The probability of obtaining k successes in n independent Bernoulli trials is given by:

P(n, k; p) = nCkpkqn – k

where p denotes the probability of success and q = (1 − p) the probability of failure.

Note: As a reminder from Chapter 5, Section 5.4, the Number of Combinations of n Objects Taken r at a Time is represented as nCk = \frac{n!}{(n-k)!k!}, where n and k are natural numbers.

 

We use the above formula to solve the following examples.

 

Example 7.1.2

If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?
Solution
Let S denote the probability of obtaining a head, and F the probability of obtaining a tail.
Clearly, n = 10, k = 3, p = 1/2, and q = 1/2.
Therefore:

Therefore: 

10C3=\frac{10!}{(10-3)!3!}=\frac{10!}{7!3!}=120

b(10, 3; 1/2) = 10C3(1/2)3(1/2)7 = 120(1/2)3(1/2)7 = .1172

 

Example 7.1.3

If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 6 out of 10 free throws in a game?
Solution
The probability of making a free throw is 3/4. Therefore, p = 3/4, q = 1/4, n = 10, and k = 6.
Therefore:
b(10, 6; 3/4) = 10C6(3/4)6(1/4)4 = .1460

 

Example 7.1.4

If a medicine cures 80% of the people who take it, what is the probability that of the eight people who take the medicine, 5 will be cured?
Solution
Here p = .80, q = .20, n = 8, and k = 5.
b(8, 5; .80) = 8C5(.80)5(.20)3 = .1468

 

Example 7.1.5

If a microchip manufacturer claims that only 4% of their chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective?
Solution
If S denotes the probability that the chip is defective, and F the probability that the chip is not defective, then p = .04, q = .96, n = 60, and k = 3.
b(60, 3; .04) = 60C3(.04)3(.96)57 = .2138

 

Example 7.1.6

If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product?
Solution
If S denoted the probability that a person will buy the product, and F the probability that the person will not buy the product, then p = .15, q = .85, n = 12, and k = 2.
b(12, 2; .15) = 12C2(.15)2(.85)10 = .2924.

 

Practice questions

1. What is the probability of getting three ones if a die is rolled five times?

2. A basketball player has an 80% chance of sinking a basket on a free throw. In five free throws, what is the probability that he will sink:

a. Only one basket?

a. Three baskets?

c. At least three baskets?

3. If a medicine cures 75% of the people who take it, what is the probability that of 30 people who take the medicine:

a. 25 will be cured?

b. 26 will be cured?

c. 27 will be cured?

d. At least 25 will be cured?

4. The Canadian Food Inspection Agency (CFIA) has found that 5% of the imported spices into Canada are contaminated with pathogenic food-borne bacteria. What is the probability that a batch of 25 imported spices will have:

a. One contaminated product?

b. Two contaminated products?

5. An executive has determined that for a door-to-door donation initiative, 20% of the households visited will provide a donation. If 10 households are visited, what is the probability that at most 2 will provide a donation?

 

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Mathematics for Public and Occupational Health Professionals Copyright © 2019 by Ian Young is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted.