Unit 18Section 18C

The Binomial Distribution

Example 1: If a coin is tossed 3 times, what is the probability of obtaining exactly 2 heads

Solution:

We can solve this problem using a tree diagram

H=0.5

T=0.5

T=0.5

T=0.5

T=0.5

T=0.5

T=0.5

H=0.5

H=0.5

H=0.5

H=0.5

H=0.5

H=0.5

T=0.5

HHH

HTH

HTT

THH

THT

TTH

TTT

HHT

There are three branches with exactly 2 heads

P(exactly 2 heads) = (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5) + (0.5)(0.5)(0.5)

= 3 (0.5)(0.5)(0.5)

= 3(0.125)

= 0.375

We can calculate this probability using a Binomial Expansion

have weh) (t expand weIf 3

302112033

3

3

2

3

1

3

0

3)( hththththt

21

2

3ht

If we examine the term

And let h= P(head) = 0.5, t=P(tail)= 0.5, we have

heads) 2exactly (375.0)5.0(3)5.0()5.0(32

3 32121 Pht

Represents the number of branches with exactly 2 heads

The Binomial Model used above applies to situations that are equivalent to drawing from a hat with replacement.

These are called BERNOULI TRIALS

Bernouli Trials have the following characteristics:

Each Trial has exactly 2 outcomes, success or failure

Each trial is independent

The probability of each outcome is the same for each trial of the experiment.

The probabilities for various events determined by a Bernouli experiment can be found easily by using the binomial expansion.

nnnnn pqn

npq

npq

npq

npq 022110 ...

210

Where p = probability of success on any trial, q = probability of failure, and p + q = 1

rrn pq

r

n termgeneral The

represents the probability of having exactly r successes in n trials if the probability of success is p.

Example 2: Determine the probability of having at least 2 girls in a family of 4 children if P(boy)=P(girl) = 0.5

40312213044

4

4

3

4

2

4

1

4

0

4gbgbgbgbgbgb

The probability of having at least 2 girls is the sum of the probabilities of having 2, 3, or 4 girls

6875.0)5.0(1)5.0(4)5.0(64

4

3

4

2

4 444403122

gbgbgb

An alternative approach considers the complementary event. The probability of having at least 2 girls is 1 – P(0 or 1 girl).

6875.0)5.0(45.011

4

0

41 441304

gbgb

A sequence of Bernouli trials has a binomial distribution. If the random variable X represents the number of successes in n trials, then

xnxqpx

nxXP

)(

Here p = the probability of success on any single trial of the experiment and p + q = 1 or q = 1 –p, hence we have

xnx ppx

nxXP

1)(

We can also express the binomial distribution in compact form, written as

X~B(n, p), read as X is distributed binomially with parameters n and p, where n is the number of trials and p = P(success)

Example 3: Write a probability function for the experiment of tossing a coin 4 times where the random variable X represents the number of heads.

Let t represent the probability of a tail and h the probability of a head.

40312213044

4

4

3

4

2

4

1

4

0

4hthththththt

Select h= 0, 1, 2, 3, 4 and calculate

0625.0)0625.0(15.00

4)5.0()5.0(

0

4

0

4)0( 40404

htXP

25.0)0625.0(45.01

4)5.0()5.0(

1

4

1

4)1( 41313

htXP

375.0)0625.0(65.02

4)5.0()5.0(

2

4

2

4)2( 42222

htXP

25.0)0625.0(45.03

4)5.0()5.0(

3

4

3

4)3( 43131

htXP

0625.0)0625.0(15.04

4)5.0()5.0(

4

4

4

4)4( 44040

htXP

So we have,

X 0 1 2 3 4

P(X=x) 0.0625 0.25 0.375 0.25 0.0625

Example 4: If X~B (7, 0.8), find P(X = 5)

Solution:

X~B (7, 0.8) means that P(X = 5) is the probability of success 5 times in 7 trials where each trial has a 0.8 chance of success and a 0.2 chance of failure.

25 2.08.05

75) P(X

So

2752512.05) P(X So

We can use the TI- 83

Binompdf(7, 0.8, 5)

Example 5: Bailey has decided that he is not going to study for the final exam in math but will instead guess at every question. If the exam consists of 50 multiple choice questions, each with 5 possible answers, what is the probability that Bailey passes the exam.

Let X represent the number of correct answers, therefore we have X~B (50, 0.2), and we want to find P(X ≥25).

P(X ≥25) = P(X = 25) + P(X = 26) + …+ P(X = 50)

This could be extremely time consuming. By using the complementary event, the calculator can help us. We will find

P(X ≥25) = 1 – P(X ≤24)

P(X ≥25) = 1 – binomcdf(50,0.2,24)

P(X ≥25) = 1 – 0.9999979051

= 0.00000209485

Bailey had better study.

P(X ≥25) = 1 – [P(X = 0) + P(X = 1) + … + P(X = 24)

Using the TI-83

Finally, it can be shown that for the binoimial distribution

X ~ B (n, p), we have

The expected value of X is µ = E(X) = np

The mode of X is the value of x which has the largest probability

The variance of X is 2 = Var(X) = npq = np( 1 – p)

The standard deviation of X is Sd(X) = npq

• HOMEWORK

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