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

The pattern in the table can help you expand any binomial by using the Binomial Theorem. Example 1B: Expanding Binomials. Use the Binomial Theorem to expand the binomial. (2 x + y ) 3. (2 x + y ) 3 = 3 C 0 (2 x ) 3 y 0 + 3 C 1 (2 x ) 2 y 1 + 3 C 2 (2 x ) 1 y 2 + 3 C 3 (2 x ) 0 y 3.

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

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

  2. 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+ 3C3(2x)0y3 = 1 • 8x3 • 1 + 3 • 4x2y + 3 • 2xy2 + 1 • 1y3 = 8x3 + 12x2y + 6xy2 + y3

  3. 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:

  4. 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.

  5. The probability that Jean will make each free throw is , or 0.5. 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? Substitute 3 for n, 1 for r, 0.5 for p, and 0.5 for q. P(r) = nCrprqn-r 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%.

  6. 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%.

  7. 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%.

  8. 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

  9. 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. 1 • List the important information: • • You make 4 trips to the drawbridge. • • The probability that the drawbridge will be down is Understand the Problem Example 3 Continued The answer will be the probability that the bridge is down at least 3 times.

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

  12. 3 Solve Example 3 Continued 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.

  13. So the probability that the drawbridge will be down for at least 3 of your trips should be greater than 4 Example 3 Continued 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.

  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.

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