11-6

1. 11-6 Binomial Distributions Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

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

3. Objectives Use the Binomial Theorem to expand a binomial raised to a power. Find binomial probabilities and test hypotheses.

4. Vocabulary Binomial Theorem binomial experiment binomial probability

5. 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.

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

7. 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)

8. 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

9. 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

10. 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:

11. 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.

12. 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%.

13. 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

14. 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?

15. 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.

16. 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."

17. 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.

18. 4 Check It Out! Example 3a Continued Look Back The answer is reasonable since it is less than but close to 1.

19. 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?

20. 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

21. 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.

22. 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.

23. 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.