Holt Algebra 2 11-6 Binomial Distributions 11-6 Binomial Distributions Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz

Holt Algebra 2

11-6Binomial Distributions11-6 Binomial Distributions

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 2

11-6Binomial 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 Algebra 2


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

Find binomial probabilities and test hypotheses.

Objectives

Holt Algebra 2


Binomial Theorembinomial experimentbinomial probability

Vocabulary

Holt Algebra 2


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 Algebra 2


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

Holt Algebra 2


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 + 5C1a4b1 + 5C2a3b2 + 5C3a2b3 +

5C4a1b4 + 5C5a0b5

= 1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5

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

Holt Algebra 2


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


Holt Algebra 2


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 Algebra 2


Check It Out! Example 1a


(x – y)5

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

Holt Algebra 2


Check It Out! Example 1b

(a + 2b)3


= a3 + 6a2b + 12ab2 + 8b3

Holt Algebra 2


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 Algebra 2


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 Algebra 2


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 Algebra 2


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 Algebra 2

11-6Binomial DistributionsCheck 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?

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

Holt Algebra 2



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?

0.2637 + 0.0879 + .0146 + 0.0010 0.3672

Holt Algebra 2


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 Algebra 2


Example 3 Continued

11 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 Algebra 2


22 Make a Plan

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

Example 3 Continued

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Solve33

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 Algebra 2


Look Back44

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 Algebra 2


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?

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

Holt Algebra 2



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?

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

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Holt Algebra 2 11-6 Binomial Distributions 11-6 Binomial Distributions Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz