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Holt Algebra 2
11-6Binomial Distributions11-6 Binomial Distributions
Holt Algebra 2
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt Algebra 2
11-6Binomial Distributions
Warm UpExpand each binomial.
1. (a + b)2 2. (x – 3y)2
Evaluate each expression.
3. 4C3 4. (0.25)0
5. 6. 23.2% of 37
a2 + 2ab + b2
x2 – 6xy + 9y2
4 1
8.584
Holt Algebra 2
11-6Binomial Distributions
Use the Binomial Theorem to expand a binomial raised to a power.
Find binomial probabilities and test hypotheses.
Objectives
Holt Algebra 2
11-6Binomial Distributions
Binomial Theorembinomial experimentbinomial probability
Vocabulary
Holt Algebra 2
11-6Binomial 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
11-6Binomial Distributions
The pattern in the table can help you expand any binomial by using the Binomial Theorem.
Holt Algebra 2
11-6Binomial 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 +
5C4a1b4 + 5C5a0b5
= 1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5
= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
Holt Algebra 2
11-6Binomial Distributions
Example 1B: Expanding Binomials
(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
Use the Binomial Theorem to expand the binomial.
Holt Algebra 2
11-6Binomial Distributions
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)
Remember!
Holt Algebra 2
11-6Binomial 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
11-6Binomial Distributions
Check It Out! Example 1b
(a + 2b)3
Use the Binomial Theorem to expand the binomial.
= a3 + 6a2b + 12ab2 + 8b3
Holt Algebra 2
11-6Binomial 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
11-6Binomial 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
11-6Binomial 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) = nCrprqn-r
P(1) = 3C1(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%.
Holt Algebra 2
11-6Binomial 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
11-6Binomial DistributionsCheck 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
11-6Binomial 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
11-6Binomial 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
11-6Binomial Distributions
Example 3 Continued
11 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
11-6Binomial Distributions
22 Make a Plan
The direct way to solve the problem is to calculate P(3) + P(4).
Example 3 Continued
Holt Algebra 2
11-6Binomial Distributions
Solve33
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.
Example 3 Continued
Holt Algebra 2
11-6Binomial Distributions
Look Back44
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
Holt Algebra 2
11-6Binomial 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
11-6Binomial 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.