Math 105, Topics in Mathematics

Introduction

6.1 Binomial Trials

Homework

Introduction

In this chapter, we talk about the number of successes in a certain number of identical and independent trials and probability.

6.1 Binomial Trials

Example 6.1.1 Suppose you toss a coin 17 times, and let X be the number of times the face H showed up. The experiment of tossing a coin is synonymous to any real life TRIAL that leads to either a SUCCESS or a FAILURE (e.g., if H showed up, we call it a success). For a unbiased coin, the probability of "success" and "failure" are same. In the case of real life trials, we are thinking of a loaded coin because in real life trials chances of success and failures need not be necessarily equal. Such a trial is called a Bernoulli trial, which has the sample space S = {s, f} where s denotes the outcome "success" and f denotes the outcome "failure." Given a Bernoulli trial, we can define a random variable Y as follows:

Y = 1	if success
Y = 0	if failure

Let us also have the notations

P(success) = p

Then P(failure) = 1-p.

So,

P(Y=1) = p

P(Y=0) = 1-p.

This random variable Y is called the Bernoulli(p) random variable.

We perform such Bernoulli trials several times, and we are interested in the number of successful outcomes. Suppose we perform n independent Bernoulli(p) trials. Let X be the number of successes in these n trials. Then X is called a Binomial(n,p) or B(n,p) random variable. X can assume values

0, 1, 2, 3, ..., n.

It may not be obvious that

P(X=r) = nCrpr(1-p) n-r

r = 0, 1, 2, ..., n

However, we do not plan to use this formula now; we plan to use some normal approximation for binomial probability.

A typical situation of binomial trials is when a pollster interviews n number of individuals to ask whether the individual would vote for a candidate or not. Then the pollster wishes to count the number of individuals X who would vote for the candidate. A trial here is interviewing one individual, and "success" means a YES answer.

Before we proceed to use normal approximation to binomial, let us note the following facts.

1. The mean of X is np and variance np(1-p).
2. We have
   

   X=X1+X2+ ... + Xn

   
   where X_r is the Bernoulli(p) variable corresponding to the rth trial, for r = 1, 2, ... , n. That is,
   

   Xr = 1	if rth trial is a success.	That means, rth voter says YES.
   Xr = 0	if rth trial is a failure.	That means, rth voter says NO.

   
   So, a B(n,p) random variable X is the sum of n Bernoulli(p) random variables.

Theorem. Suppose X is B(n,p) random variable. If n is large and p is not too close to 0 or 1, then X is approximately normally distributed with mean μ = np and standard deviation σ = (np(1-p))1/2.

This is the theorem we use to compute binomial probability. Please keep in mind that we get only an approximate probability, and it works well enough if n is large enough.

Problems on 6.1: Binomial Trials

Exercise 6.1.1. A Lawrence bank knows that 35 percent (i.e., p=0.35) of the customers visit the driveway counters. If 400 customers visit the bank, what is the approximate probability that more than 150 visit the driveway counter?

Solution: Suppose X is the number of customers out of 400 who would visit the driveway counter. The X is B(400, 0.35)-variable. The means X is

μ = np = 400x0.35 = 140
and the standard deviation
σ = (np(1-p))^1/2 = (400 x0.35 x 0.65) ^1/2 = 9.5394.

We compute approximate probability
P(150 < X ) = P(151 ≤ X) =
P((151- μ)/σ < (X- μ )/σ) = P((151-140)/9.5394 ≤ Z ) =
P(1.15 ≤ Z) =
1 - P(Z < 1.15) = 1- 0.8749 = 0.1251.

Look at the flash animated Solution.

Exercise 6.1.2. It is known that the probability that a household owns a pressure cooker is p = 0.1. If 190 households are interviewed, find the approximate probability that

1. at least 25 households own a pressure cooker, and
2. at most 30 households own a pressure cooker.

Solution:

1. Here n = 190, and p = 0.1. Suppose X is the number of households among these 190 who own a pressure cooker. We are asked to find P(X > 25).
   
   
2. The mean of X> = μ = 0.1x190 = 19.
   
   
3. The standard deviation X = σ = (np(1-p))^1/2 = (190 x 0.1 x 0.9) ^1/2 = 4.1352.
   
   
4. So we compute approximate probability
   P( 25 ≤ X ) =
   P((25-μ)/σ ≤ (X - μ)/σ) =
   P((25-19)/4.1352 < Z ) =
   P(1.45 < Z) =
   1 - P(Z < 1.45) = 1- 0.9265 = 0.0735.
   
   
5. Now we compute approximate probability P(X ≤ 30).
   We have
   P(X < 30) =
   P((X- μ)/σ < (30- μ)/σ =
   P(Z < (30 - 19)/4.1352) =
   P(Z < 2.66) =0.9961.
   
   
6. Look at the flash-animated Solution.

Exercise 6.1.3. The campaign committee of a candidate claims that sixty percent (p = 0.6) of the voters are in favor of the candidate. You interview 150 voters. Assuming the campaign committee's claim, find what is the approximate probability that at most 87 will favor the candidate?

Solution: Here n = 150, and p = 0.6. Suppose X is the number of voters among these 150 who are in favor of this candidate. Then X is a B(n,p) = B(150,0.6) - variable.

1. The mean of X is
   µ = np = 150 x 0.6 = 90.
   
   
2. The standard deviation of X
   σ = (np(1-p))^1/2 = (150 x 0.6 x 0.4) ^1/2 = 6.
   
   
3. We compute approximate probability
   P(X ≤ 87) =
   P((X - μ)/σ < (87 - μ)/ σ) =
   P(Z < (87 - 90)/6) =
   P(Z < -0.5) = 0.3085.

Look at the flash-animated Solution.

Exercise 6.1.4. A technique is used to fertilize eggs in a fertility clinic laboratory. It is known that the probability that an egg will be fertilized by this technique is p = 0.1. If 500 eggs are treated, then what is the probability that at least 60 eggs will be fertilized?

Solution: Here n = 500 and p = 0.1. Suppose X is the number of eggs out of these 500 eggs that got fertilized. Then X is B(n,p) = B(500,0.1)-variable. We are asked to find approximate P(X ≤ 60).

1. The mean of X
   μ = np = 500 x 0.1 =50
   
   
2. The standard deviation
   σ = (np(1-p))^1/2 = (500 x 0.1 x 0.9) ^1/2 = 6.7082.
   
   
3. We have
   P(60 ≤ X ) =
   P((60 - μ)/σ ≤ (X - μ)/σ) =
   P((60 - 50)/6.7082 < Z) =
   P(1.49 < Z) = 1 - P(Z < 1.49) = 1- 0.9319 = 0.0681.

Examine the flash-animated Solution.

Exercise 6.1.5. The probability that a computer chip produced in a factory is defective is p = 0.2. If you have a sample of 60 chips, what is the probability that the number of defective will be at most 19?

Solution: Here n= 60, and p = 0.2. Suppose X is the number of defective items in these 60 chips. We asked to compute approximate P(X < 20).

1. The mean of X is µ = np =60 x 0.2 = 12
   
   
2. The standard deviation of X
   σ = (np(1-p))^1/2 = (60 x 0.2 x 0.8) ^1/2 = 3.0984.
   
   
3. P(X ≤ 19) =
   P((X- μ)/σ ≤ (19- μ)/σ) =
   P(Z < (19 - 12)/3.0984) =
   P(Z < 2.26) = 0.9881.

Examine the flash-animated Solution.

Exercise 6.1.6. The probability that a light bulb produced by a machine is defective is p = 0.2. Suppose a quality control inspector takes a sample of 120 bulbs. What is the probability that more than 30 bulbs will be defective?
Solution

Exercise 6.1.7. Suppose the probability that a student has access to the Internet is p = 0.8. Suppose you interview 160 students. What is the probability that less than 120 students will have access to the Internet?
Solution

Exercise 6.1.8. Suppose that the probability that a person favors medical use of marijuana is p = 0.6. If 780 individuals are interviewed, what is the probability that less than 450 will be in favor?
Solution

Exercise 6.1.9. Suppose the probability that a middle-income family invests in the stock market is p = 0.8. If we interview 880 middle-income families, then what is the probability that more than 700 would have invested in stock market?
Solution

Exercise 6.1.10. Suppose an insurance company knows from experience that the probability that a life insurance policy holder will survive another ten years is p = 0.9. The company has 2280 policies. What is the probability that more than 2025 will survive another ten years? Solution