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# 11-6 - PowerPoint PPT Presentation

11-6. Binomial Distributions. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. Warm Up Expand each binomial. 1. ( a + b ) 2 2. ( x – 3 y ) 2 Evaluate each expression. 3. 4 C 3 4. (0.25) 0 5. 6. 23.2% of 37. x 2 – 6 xy + 9 y 2. a 2 + 2 ab + b 2. 1. 4. 8.584.

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11-6

Binomial Distributions

Warm Up

Lesson Presentation

Lesson Quiz

Holt Algebra 2

Warm Up

Expand each binomial.

1. (a + b)22. (x – 3y)2

Evaluate each expression.

3.4C34. (0.25)0

5.6. 23.2% of 37

x2 – 6xy + 9y2

a2 + 2ab + b2

1

4

8.584

Objectives

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

Find binomial probabilities and test hypotheses.

Vocabulary

Binomial Theorem

binomial experiment

binomial probability

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.

Remember!

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

Check It Out! Example 1a

Use the Binomial Theorem to expand the binomial.

(x – y)5

(x – y)5 = 5C0x5(–y)0 + 5C1x4(–y)1+ 5C2x3(–y)2+ 5C3x2(–y)3+5C4x1(–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

Check It Out! Example 1b

Use the Binomial Theorem to expand the binomial.

(a + 2b)3

(a + 2b)3 = 3C0a3(2b)0 + 3C1a2(2b)1 + 3C2a1(2b)2 + 3C3a0(2b)3

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

= a3 + 6a2b + 12ab2 + 8b3

A binomial experimentconsists 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:

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.

The probability that the counselor will be assigned 1 of the 3 students is .

Substitute 3 for n, 2 for r,

for p, and for q.

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?

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

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?

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

The probability of answering a question correctly is 0.25.

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

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

0.2637 + 0.0879 + .0146 + 0.0010  0.3672

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?

1

Understand the Problem

• List the important information:
• • Twenty questions with four choices
• • The probability of guessing a correct answer is .

Check It Out! Example 3a Continued

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

Make a Plan

2

Check It Out! Example 3a Continued

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

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

3

Solve

Check It Out! Example 3a Continued

Step 1 Find P(0 or 1 correct).

P(0) + P(1)

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

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

 0.0032+ 0.0211

 0.0243

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.

4

Check It Out! Example 3a Continued

Look Back

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

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?

1

Understand the Problem

Check It Out! Example 3b Continued

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

Make a Plan

2

Check It Out! Example 3b Continued

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

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.

3

Solve

Check It Out! Example 3b Continued

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

P(24) + P(25)

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

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

 0.3079 + 0.6035

 0.9114

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.

4

Check It Out! Example 3b Continued

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.