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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 Up Expand 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 = 5C0a 5b0 + 5C1a

4b1 + 5C2a 3b2 + 5C3a

2b3 +

5C4a 1b4 + 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 = 5C0x 5(–y)0 + 5C1x

4(–y)1 + 5C2x 3(–y)2 +

5C3x 2(–y)3 + 5C4x

1(–y)4 + 5C5x 0(–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 = 3C0a 3(2b)0 + 3C1a

2(2b)1 + 3C2a 1(2b)2 +

3C3a 0(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) = nCrp rqn-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