# 7.1. Binomial Probability

# 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:

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

- If a coin is flipped 10 times, what is the probability that it will fall heads 3 times?
- 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?
- 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?
- 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?
- 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.**four hits

**b.**three hits

**c.**two hits

**d.**one hit

**e.**no hits.

**Solution**

*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.

*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 }

^{2}(0.7)

^{2}, the probability of obtaining two successes is 6(0.3)

^{2}(0.7)

^{2}.

*S*and three

*F*‘s, there are four such outcomes: SFFF , FSFF , FFSF , and FFFS.

^{3}, the probability of getting one hit is 4(0.3)(0.7)

^{3}.

*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*) = *n*C*kp*^{k}*q*^{n – k}

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

We use the above formula to solve the following examples.

Example 7.1.2

**Solution**

*S*denote the probability of obtaining a head, and

*F*the probability of obtaining a tail.

*n*= 10,

*k*= 3,

*p*= 1/2, and

*q*= 1/2.

*b*(10, 3; 1/2) = 10C3(1/2)

^{3}(1/2)

^{7}= .1172

Example 7.1.3

**Solution**

*p*= 3/4,

*q*= 1/4,

*n*= 10, and

*k*= 6.

*b*(10, 6; 3/4) = 10C6(3/4)

^{6}(1/4)

^{4}= .1460

Example 7.1.4

**Solution**

*p*= .80,

*q*= .20,

*n*= 8, and

*k*= 5.

*b*(8, 5; .80) = 8C5(.80)

^{5}(.20)

^{3}= .1468

Example 7.1.5

**Solution**

*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

**Solution**

*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.

^{2}(.85)

^{10}= .2924.

# Practice questions

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

**2. **If a medicine cures 75% of the people who take it, what is the probability that of 30 people who take the medicine, 25 will be cured?

**3.** A basketball player has an 80% chance of sinking a basket on a free throw. What is the probability that he will sink *at least three* baskets in five free throws?

**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 two contaminated products?

Answers

**1.** 0.0322

**2.** 0.1047

**3.** 0.9421

**4.** 0.2305