Binomial Distributions How do we use the Binomial

Binomial Distributions • How do we use the Binomial Theorem to expand a binomial raised to a power? • How do we find binomial probabilities and test hypotheses? Holt Mc. Dougal Algebra 2

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 Mc. Dougal Algebra 2

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 3 C 2 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 Mc. Dougal Algebra 2

Binomial Distributions Example 1: 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) = n. Cr prqn-r Substitute 3 for n, 1 for r, 0. 5 for p, and 0. 5 for q. P(1) = 3 C 1 (0. 5)3 -1 The probability that Jean will make exactly one free throw is 37. 5%. Holt Mc. Dougal Algebra 2

Binomial Distributions Example 2: 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) 1(0. 5)3 -1 + C (0. 5)2(0. 5)3 -2 + C (0. 5)3 -3 C (0. 5) 3 2 3 3 3 1 The probability that Jean will make at least one free throw is 87. 5%. Holt Mc. Dougal Algebra 2

Binomial Distributions Example 3: Finding Binomial Probabilities 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 the counselor will be assigned 1 of the 3 students is. Substitute 3 for n, 2 for r, for p, and for q. The probability that Counselor Jenkins will get 2 of the next 3 students assigned is about 22%. Holt Mc. Dougal Algebra 2

Binomial Distributions Example 4: Finding Binomial Probabilities 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) 2(0. 75)5 -2 + C (0. 25)3(0. 75)5 -3 + C (0. 25)4(0. 75)5 -4 C (0. 25) 5 2 5 4 5 3 + 5 C 5(0. 25)5(0. 75)5 -5 Holt Mc. Dougal Algebra 2

Binomial Distributions Lesson 3. 3 Practice B Holt Mc. Dougal Algebra 2