11-6 11-6Binomial BinomialDistributions Distributions Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 22 11-6 Binomial Distributions Warm Up Expand each binomial. x2 – 6xy + 9y2 1. (a + b)2 a2 + 2ab + b2 2. (x – 3y)2 Evaluate each expression. 3. 4C3 4 4. (0.25)0 1 5. 6. 23.2% of 37 8.584 Holt Algebra 2 11-6 Binomial Distributions Objectives Use the Binomial Theorem to expand a binomial raised to a power. Find binomial probabilities and test hypotheses. Holt Algebra 2 11-6 Binomial Distributions Vocabulary Binomial Theorem binomial experiment binomial probability Holt Algebra 2 11-6 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 Algebra 2 11-6 Binomial Distributions The pattern in the table can help you expand any binomial by using the Binomial Theorem. Holt Algebra 2 11-6 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 = 5C0a5b0 + 5C1a4b1 + 5C2a3b2 + 5C3a2b3 + 1b4 + C a0b5 C a 5 4 5 5 = 1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Holt Algebra 2 11-6 Binomial Distributions Example 1B: Expanding Binomials Use the Binomial Theorem to expand the binomial. (2x + y)3 (2x + y)3 = 3C0(2x)3y0 + 3C1(2x)2y1 + 3C2(2x)1y2 + 0y3 C (2x) 3 3 = 1 • 8x3 • 1 + 3 • 4x2y + 3 • 2xy2 + 1 • 1y3 = 8x3 + 12x2y + 6xy2 + y3 Holt Algebra 2 11-6 Binomial Distributions Remember! 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) Holt Algebra 2 11-6 Binomial Distributions Check It Out! Example 1a Use the Binomial Theorem to expand the binomial. (x – y)5 = x5 – 5x4y + 10x3y2 – 10x2y3 + 5xy4 – y5 Holt Algebra 2 11-6 Binomial Distributions Check It Out! Example 1b Use the Binomial Theorem to expand the binomial. (a + 2b)3 = a3 + 6a2b + 12ab2 + 8b3 Holt Algebra 2 11-6 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 Algebra 2 11-6 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 Algebra 2 11-6 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 Substitute 3 for n, 1 for r, 0.5 for p, and 0.5 for q. P(1) = 3C1(0.5)1(0.5)3-1 = 3(0.5)(0.25) = 0.375 The probability that Jean will make exactly one free throw is 37.5%. Holt Algebra 2 11-6 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 Algebra 2 11-6 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? The probability that Counselor Jenkins will get 2 of the next 3 students assigned is about 22%. Holt Algebra 2 11-6 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? 0.2637 + 0.0879 + .0146 + 0.0010 0.3672 Holt Algebra 2 11-6 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 Algebra 2 11-6 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 Algebra 2 11-6 Binomial Distributions Example 3 Continued 2 Make a Plan The direct way to solve the problem is to calculate P(3) + P(4). Holt Algebra 2 11-6 Binomial Distributions Example 3 Continued 3 Solve 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. Holt Algebra 2 11-6 Binomial Distributions Example 3 Continued 4 Look Back 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 11-6 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? The probability that Wendy will get at least 2 answers correct is about 0.98. Holt Algebra 2 11-6 Binomial Distributions 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? The probability that there are 23 or fewer acceptable parts is about 0.09. Holt Algebra 2