1. Holt McDougal Algebra 2 Binomial Distributions The pattern in the table can help you expand any binomial by using the Binomial Theorem.

2. Holt McDougal Algebra 2 Binomial Distributions Example 1B: Expanding Binomials (2x + y) 3 (2x + y) 3 = 3 C 0 (2x) 3 y 0 + 3 C 1 (2x) 2 y 1 + 3 C 2 (2x) 1 y 2 + 3 C 3 (2x) 0 y 3 = 1 8x 3 1 + 3 4x 2 y + 3 2xy 2 + 1 1y 3 = 8x 3 + 12x 2 y + 6xy 2 + y 3 Use the Binomial Theorem to expand the binomial.

3. Holt McDougal 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:

4. Holt McDougal 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.

5. Holt McDougal Algebra 2 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) = n C r p r q n-r P(1) = 3 C 1 (0.5) 1 (0.5) 3-1 Substitute 3 for n, 1 for r, 0.5 for p, and 0.5 for q. = 3(0.5)(0.25) = 0.375 The probability that Jean will make exactly one free throw is 37.5%.

6. Holt McDougal Algebra 2 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 + 3 C 2 (0.5) 2 (0.5) 3-2 + 3 C 3 (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%.

7. Holt McDougal Algebra 2 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? 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%. The probability that the counselor will be assigned 1 of the 3 students is.

8. Holt McDougal Algebra 2 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? The probability of answering a question correctly is 0.25. 5 C 2 (0.25) 2 (0.75) 5-2 + 5 C 3 (0.25) 3 (0.75) 5-3 + 5 C 4 (0.25) 4 (0.75) 5-4 + 5 C 5 (0.25) 5 (0.75) 5-5 At least 2 answers correct is the same as exactly 2, 3, 4, or 5 questions correct. P(2) + P(3) + P(4) + P(5) 0.2637 + 0.0879 +.0146 + 0.0010  0.3672

9. Holt McDougal Algebra 2 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?

10. Holt McDougal Algebra 2 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

11. Holt McDougal Algebra 2 Binomial Distributions 2 Make a Plan The direct way to solve the problem is to calculate P(3) + P(4). Example 3 Continued

12. Holt McDougal Algebra 2 Binomial Distributions Solve 3 P(3) + P(4) = 4 C 3 (0.80) 3 (0.20) 4-3 + 4 C 4 (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. Example 3 Continued

13. Holt McDougal Algebra 2 Binomial Distributions Look Back 4 Example 3 Continued 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

14. Holt McDougal Algebra 2 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?

15. Holt McDougal Algebra 2 Binomial Distributions Check It Out! Example 3a Continued 1 Understand the Problem The answer will be the probability she will get at least 2 answers correct by guessing. List the important information: Twenty questions with four choices The probability of guessing a correct answer is.

16. Holt McDougal Algebra 2 Binomial Distributions 2 Make a Plan The direct way to solve the problem is to calculate P(2) + P(3) + P(4) + … + P(20). Check It Out! Example 3a Continued An easier way is to use the complement. "Getting 0 or 1 correct" is the complement of "getting at least 2 correct."

17. Holt McDougal Algebra 2 Binomial Distributions Solve 3 = 20 C 0 (0.25) 0 (0.75) 20-0 + 20 C 1 (0.25) 1 (0.75) 20-1 Check It Out! Example 3a Continued P(0) + P(1) = 1(0.25) 0 (0.75) 20 + 20(0.25) 1 (0.75) 19  0.0032 + 0.0211  0.0243 Step 1 Find P(0 or 1 correct). 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. Holt McDougal Algebra 2 Binomial Distributions Check It Out! Example 3a Continued Look Back 4 The answer is reasonable since it is less than but close to 1.

19. Holt McDougal Algebra 2 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?

20. Holt McDougal Algebra 2 Binomial Distributions Check It Out! Example 3b Continued 1 Understand the Problem 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. Holt McDougal Algebra 2 Binomial Distributions 2 Make a Plan The direct way to solve the problem is to calculate P(1) + P(2) + P(3) + … + P(23). Check It Out! Example 3b Continued 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. Holt McDougal Algebra 2 Binomial Distributions Solve 3 = 25 C 24 (0.98) 24 (0.02) 25-24 + 25 C 25 (0.98) 25 (0.02) 25-25 Check It Out! Example 3b Continued P(24) + P(25) Step 1 Find P(24 or 25 acceptable parts). = 25(0.98) 24 (0.02) 1 + 1(0.98) 25 (0.02) 0  0.3079 + 0.6035 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.  0.9114

23. Holt McDougal Algebra 2 Binomial Distributions 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. 4 Check It Out! Example 3b Continued