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Holt McDougal Algebra 2
Binomial DistributionsBinomial Distributions
Holt Algebra 2
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt McDougal Algebra 2
Binomial Distributions
Warm Up Expand 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 McDougal Algebra 2
Binomial Distributions
Use the Binomial Theorem to expand a binomial raised to a power.
Find binomial probabilities and test hypotheses.
Objectives
Holt McDougal Algebra 2
Binomial Distributions
Binomial Theorem
binomial experiment
binomial probability
Vocabulary
Holt McDougal Algebra 2
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 McDougal Algebra 2
Binomial Distributions
The pattern in the table can help you expand any binomial by using the Binomial Theorem.
Holt McDougal Algebra 2
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 = 5C0a 5b0 + 5C1a
4b1 + 5C2a 3b2 + 5C3a
2b3 +
5C4a 1b4 + 5C5a
0b5
= 1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 +
1a0b5
= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
Holt McDougal Algebra 2
Binomial 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 McDougal Algebra 2
Binomial 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 McDougal Algebra 2
Binomial Distributions
Check It Out! Example 1a
Use the Binomial Theorem to expand the binomial.
(x – y)5
(x – y)5 = 5C0x 5(–y)0 + 5C1x
4(–y)1 + 5C2x 3(–y)2 +
5C3x 2(–y)3 + 5C4x
1(–y)4 + 5C5x 0(–y)5
= 1x5(–y)0 + 5x4(–y)1 + 10x3(–y)2 + 10x2(–y)3 + 5x1(–y)4 + 1x0(–y)5
= x5 – 5x4y + 10x3y2 – 10x2y3 + 5xy4 – y5
Holt McDougal Algebra 2
Binomial Distributions
Check It Out! Example 1b
(a + 2b)3
(a + 2b)3 = 3C0a 3(2b)0 + 3C1a
2(2b)1 + 3C2a 1(2b)2 +
3C3a 0(2b)3
= 1 • a3 • 1 + 3 • a2 • 2b + 3 • a • 4b2 + 1 • 1 • 8b3
= a3 + 6a2b + 12ab2 + 8b3
Use the Binomial Theorem to expand the binomial.
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:
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 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 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) = nCrp rqn-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 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 + 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 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 .
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.
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
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
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?
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
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
Holt McDougal Algebra 2
Binomial Distributions
Solve3
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 McDougal Algebra 2
Binomial Distributions
Look Back4
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 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?
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 .
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."
Holt McDougal Algebra 2
Binomial Distributions
Solve3
= 20C0(0.25) 0(0.75)20-0 + 20C1(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
