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11-6

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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.

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

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.