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Holt McDougal Algebra 2

Binomial DistributionsBinomial Distributions

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz

Holt McDougal Algebra 2

Holt McDougal Algebra 2

Binomial Distributions

Warm UpExpand each binomial.

1. (a + b)2 2. (x – 3y)2

Evaluate each expression.

3. 4C3 4. (0.25)0

5. 6. 23.2% of 37

a2 + 2ab + b2

x2 – 6xy + 9y2

4 1

8.584

Holt McDougal Algebra 2

Binomial Distributions

Use the Binomial Theorem to expand a binomial raised to a power.

Find binomial probabilities and test hypotheses.

Objectives

Holt McDougal Algebra 2

Binomial Distributions

Binomial Theorem

binomial experiment

binomial probability

Vocabulary

Holt McDougal Algebra 2

Binomial Distributions

You used Pascal’s triangle to find binomial expansions in Lesson 6-2. The coefficients of the expansion of(x + y)n are the numbers in Pascal’s triangle, which are actually combinations.

Holt McDougal Algebra 2

Binomial Distributions

The pattern in the table can help you expand any binomial by using the Binomial Theorem.

Holt McDougal Algebra 2

Binomial Distributions

Example 1A: Expanding Binomials

Use the Binomial Theorem to expand the binomial.

(a + b)5 The sum of the exponents for each term is 5.

(a + b)5 = 5C0a5b0 + 5C1a

4b1 + 5C2a3b2 + 5C3a

2b3 +

5C4a1b4 + 5C5a

0b5

= 1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 +

1a0b5

= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Holt McDougal Algebra 2

Binomial Distributions

Example 1B: Expanding Binomials

(2x + y)3

(2x + y)3 = 3C0(2x)3y0 + 3C1(2x)2y1 + 3C2(2x)1y2 +

3C3(2x)0y3

= 1 • 8x3 • 1 + 3 • 4x2y + 3 • 2xy2 + 1 • 1y3

= 8x3 + 12x2y + 6xy2 + y3

Use the Binomial Theorem to expand the binomial.

Holt McDougal Algebra 2

Binomial Distributions

In the expansion of (x + y)n, the powers of x decrease from n to 0 and the powers of y increase from 0 to n. Also, the sum of the exponents is n for each term. (Lesson 6-2)

Remember!

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 1a

Use the Binomial Theorem to expand the binomial.

(x – y)5

(x – y)5 = 5C0x5(–y)0 + 5C1x

4(–y)1 + 5C2x3(–y)2 +

5C3x2(–y)3 + 5C4x

1(–y)4 + 5C5x0(–y)5

= 1x5(–y)0 + 5x4(–y)1 + 10x3(–y)2 + 10x2(–y)3 + 5x1(–y)4 + 1x0(–y)5

= x5 – 5x4y + 10x3y2 – 10x2y3 + 5xy4 – y5

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 1b

(a + 2b)3

(a + 2b)3 = 3C0a3(2b)0 + 3C1a

2(2b)1 + 3C2a1(2b)2 +

3C3a0(2b)3

= 1 • a3 • 1 + 3 • a2 • 2b + 3 • a • 4b2 + 1 • 1 • 8b3

= a3 + 6a2b + 12ab2 + 8b3

Use the Binomial Theorem to expand the binomial.

Holt McDougal Algebra 2

Binomial Distributions

A binomial experiment consists of n independent trials whose outcomes are either successes or failures; the probability of success p is the same for each trial, and the probability of failure q is the same for each trial. Because there are only two outcomes, p + q = 1, or q = 1 - p. Below are some examples of binomial experiments:

Holt McDougal Algebra 2

Binomial Distributions

Suppose the probability of being left-handed is 0.1 and you want to find the probability that 2 out of 3 people will be left-handed. There are 3C2 ways to choose the two left-handed people: LLR, LRL, and RLL. The probability of each of these occurring is 0.1(0.1)(0.9). This leads to the following formula.

Holt McDougal Algebra 2

Binomial Distributions

Example 2A: Finding Binomial Probabilities

Jean usually makes half of her free throws in basketball practice. Today, she tries 3 free throws. What is the probability that Jean will make exactly 1 of her free throws?

The probability that Jean will make each free throw is

, or 0.5.

P(r) = nCrprqn-r

P(1) = 3C1(0.5)1(0.5)3-1

Substitute 3 for n, 1 for r,0.5 for p, and 0.5 for q.

= 3(0.5)(0.25) = 0.375

The probability that Jean will make exactly one free throw is 37.5%.

Holt McDougal Algebra 2

Binomial Distributions

Example 2B: Finding Binomial Probabilities

Jean usually makes half of her free throws in basketball practice. Today, she tries 3 free throws. What is the probability that she will make at least 1 free throw?

At least 1 free throw made is the same as exactly 1, 2, or 3 free throws made.

P(1) + P(2) + P(3)

0.375 + 3C2(0.5)2(0.5)3-2 + 3C3(0.5)3(0.5)3-3

0.375 + 0.375 + 0.125 = 0.875

The probability that Jean will make at least one free throw is 87.5%.

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 2a

Students are assigned randomly to 1 of 3 guidance counselors. What is the probability that Counselor Jenkins will get 2 of the next 3 students assigned?

Substitute 3 for n, 2 for r,

for p, and for q.

The probability that Counselor Jenkins will get 2 of the next 3 students assigned is about 22%.

The probability that the counselor will be assigned 1

of the 3 students is .

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 2b

Ellen takes a multiple-choice quiz that has 5 questions, with 4 answer choices for each question. What is the probability that she will get at least 2 answers correct by guessing?

The probability of answering a question correctly is 0.25.

5C2(0.25)2(0.75)5-2 + 5C3(0.25)3(0.75)5-3 +

5C4(0.25)4(0.75)5-4 + 5C5(0.25)5(0.75)5-5

At least 2 answers correct is the same as exactly 2, 3, 4, or 5 questions correct.

P(2) + P(3) + P(4) + P(5)

0.2637 + 0.0879 + .0146 + 0.0010 0.3672

Holt McDougal Algebra 2

Binomial Distributions

Example 3: Problem-Solving Application

You make 4 trips to a drawbridge. There is a 1 in 5 chance that the drawbridge will be raiseD when you arrive. What is the probability that the bridge will be down for at least 3 of your trips?

Holt McDougal Algebra 2

Binomial Distributions

Example 3 Continued

1 Understand the Problem

The answer will be the probability that the bridge is down at least 3 times.

List the important information:

• You make 4 trips to the drawbridge.

• The probability that the drawbridge will

be down is

Holt McDougal Algebra 2

Binomial Distributions

2 Make a Plan

The direct way to solve the problem is to calculate P(3) + P(4).

Example 3 Continued

Holt McDougal Algebra 2

Binomial Distributions

Solve3

P(3) + P(4)

= 4C3(0.80)3(0.20)4-3 + 4C4(0.80)4(0.20)4-3

= 4(0.80)3(0.20) + 1(0.80)4(1)

= 0.4096 + 0.4096

= 0.8192

The probability that the bridge will be down for at least 3 of your trips is 0.8192.

Example 3 Continued

Holt McDougal Algebra 2

Binomial Distributions

Look Back4

Example 3 Continued

The answer is reasonable, as the expected number of trips the drawbridge will be down is of 4, = 3.2, which is greater than 3.

So the probability that the drawbridge will be

down for at least 3 of your trips should be

greater than

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 3a

Wendy takes a multiple-choice quiz that has 20 questions. There are 4 answer choices for each question. What is the probability that she will get at least 2 answers correct by guessing?

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 3a Continued

1 Understand the Problem

The answer will be the probability she will get at least 2 answers correct by guessing.

List the important information:

• Twenty questions with four choices

• The probability of guessing a correct answer is .

Holt McDougal Algebra 2

Binomial Distributions

2 Make a Plan

The direct way to solve the problem is to calculate P(2) + P(3) + P(4) + … + P(20).

Check It Out! Example 3a Continued

An easier way is to use the complement. "Getting 0 or 1 correct" is the complement of "getting at least 2 correct."

Holt McDougal Algebra 2

Binomial Distributions

Solve3

= 20C0(0.25)0(0.75)20-0 + 20C1(0.25)1(0.75)20-1

Check It Out! Example 3a Continued

P(0) + P(1)

= 1(0.25)0(0.75)20 + 20(0.25)1(0.75)19

0.0032 + 0.0211

0.0243

Step 1 Find P(0 or 1 correct).

Step 2 Use the complement to find the probability.

1 – 0.0243 0.9757

The probability that Wendy will get at least 2 answers correct is about 0.98.

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 3a Continued

Look Back4

The answer is reasonable since it is less than but close to 1.

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 3b

A machine has a 98% probability of producing a part within acceptable tolerance levels. The machine makes 25 parts an hour. What is the probability that there are 23 or fewer acceptable parts?

Holt McDougal Algebra 2

Binomial Distributions

Check It Out! Example 3b Continued

1 Understand the Problem

The answer will be the probability of getting 1–23 acceptable parts.

List the important information:

• 98% probability of an acceptable part

• 25 parts per hour with 1–23 acceptable parts

Holt McDougal Algebra 2

Binomial Distributions

2 Make a Plan

The direct way to solve the problem is to calculate P(1) + P(2) + P(3) + … + P(23).

Check It Out! Example 3b Continued

An easier way is to use the complement. "Getting 23 or fewer" is the complement of "getting greater than 23.“ Find this probability, and then subtract the result from 1.

Holt McDougal Algebra 2

Binomial Distributions

Solve3

= 25C24(0.98)24(0.02)25-24 + 25C25(0.98)25(0.02)25-25

Check It Out! Example 3b Continued

P(24) + P(25)

Step 1 Find P(24 or 25 acceptable parts).

= 25(0.98)24(0.02)1 + 1(0.98)25(0.02)0

0.3079 + 0.6035

Step 2 Use the complement to find the probability.

1 – 0.9114 0.0886

The probability that there are 23 or fewer acceptable parts is about 0.09.

0.9114

Holt McDougal Algebra 2

Binomial Distributions

Look Back

Since there is a 98% chance that a part will be produced within acceptable tolerance levels, the probability of 0.09 that 23 or fewer acceptable parts are produced is reasonable.

4

Check It Out! Example 3b Continued

Holt McDougal Algebra 2

Binomial Distributions

Lesson Quiz: Part I

Use the Binomial Theorem to expand each binomial.

1. (x + 2)4

2. (2a – b)5

A binomial experiment has 4 trials, with p = 0.3.

3. What is the probability of 1 success?

4. What is the probability of at least 2 successes?

x4 + 8x3 + 24x2 + 32x + 16

32a5 – 80a4b + 80a3b2 – 40a2b3 + 10ab4 – b5

0.4116

0.3483

Holt McDougal Algebra 2

Binomial Distributions

Lesson Quiz: Part II

A binomial experiment has 4 trials, with p = 0.3.

5. There is a 10% chance that Nila will have to wait for a train to pass as she heads for school. What is the probability that she will not have to wait for a train all 5 days this week?

6. Krissy has 3 arrows. The probability of her hitting

the target is . What is the probability that she will

get at least one arrow on the target?

about 59%

78.4%