Transcript Slide 1

11-6
11-6Binomial
BinomialDistributions
Distributions
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
11-6 Binomial Distributions
Warm Up
Expand each binomial.
x2 – 6xy + 9y2
1. (a + b)2 a2 + 2ab + b2 2. (x – 3y)2
Evaluate each expression.
3. 4C3 4
4. (0.25)0 1
5.
6. 23.2% of 37
8.584
Holt Algebra 2
11-6 Binomial Distributions
Objectives
Use the Binomial Theorem to expand a
binomial raised to a power.
Find binomial probabilities and test
hypotheses.
Holt Algebra 2
11-6 Binomial Distributions
Vocabulary
Binomial Theorem
binomial experiment
binomial probability
Holt Algebra 2
11-6 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 Algebra 2
11-6 Binomial Distributions
The pattern in the table can help you expand any
binomial by using the Binomial Theorem.
Holt Algebra 2
11-6 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 = 5C0a5b0 + 5C1a4b1 + 5C2a3b2 + 5C3a2b3 +
1b4 + C a0b5
C
a
5 4
5 5
= 1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 +
1a0b5
= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
Holt Algebra 2
11-6 Binomial Distributions
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 +
0y3
C
(2x)
3 3
= 1 • 8x3 • 1 + 3 • 4x2y + 3 • 2xy2 + 1 • 1y3
= 8x3 + 12x2y + 6xy2 + y3
Holt Algebra 2
11-6 Binomial Distributions
Remember!
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)
Holt Algebra 2
11-6 Binomial Distributions
Check It Out! Example 1a
Use the Binomial Theorem to expand the
binomial.
(x – y)5
(x – y)5 = 5C0x5(–y)0 + 5C1x4(–y)1 + 5C2x3(–y)2 +
2
3
1
4
0
5
5C3x (–y) + 5C4x (–y) + 5C5x (–y)
= 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 Algebra 2
11-6 Binomial Distributions
Check It Out! Example 1b
Use the Binomial Theorem to expand the
binomial.
(a + 2b)3
(a + 2b)3 = 3C0a3(2b)0 + 3C1a2(2b)1 + 3C2a1(2b)2 +
0
3
3C3a (2b)
= 1 • a3 • 1 + 3 • a2 • 2b + 3 • a • 4b2 +
1 • 1 • 8b3
= a3 + 6a2b + 12ab2 + 8b3
Holt Algebra 2
11-6 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 Algebra 2
11-6 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 Algebra 2
11-6 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
Substitute 3 for n, 1 for r,
0.5 for p, and 0.5 for q.
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%.
Holt Algebra 2
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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%.
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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 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%.
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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 +
4
5-4 + C (0.25)5(0.75)5-5
5C4(0.25) (0.75)
5 5
0.2637 + 0.0879 + .0146 + 0.0010  0.3672
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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
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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
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Example 3 Continued
2
Make a Plan
The direct way to solve the problem is to
calculate P(3) + P(4).
Holt Algebra 2
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Example 3 Continued
3
Solve
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.
Holt Algebra 2
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Example 3 Continued
4
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.
So the probability that the drawbridge will be
down for at least 3 of your trips should be
greater than
Holt Algebra 2
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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 Algebra 2
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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 Algebra 2
.
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Check It Out! Example 3a Continued
2
Make a Plan
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."
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Check It Out! Example 3a Continued
3
Solve
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.
Holt Algebra 2
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Check It Out! Example 3a Continued
4
Look Back
The answer is reasonable since it is less than
but close to 1.
Holt Algebra 2
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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?
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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
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Check It Out! Example 3b Continued
2
Make a Plan
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.
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Check It Out! Example 3b Continued
3
Solve
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.
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Check It Out! Example 3b Continued
4
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.
Holt Algebra 2
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Lesson Quiz: Part I
Use the Binomial Theorem to expand each
binomial.
1. (x + 2)4
x4 + 8x3 + 24x2 + 32x + 16
5 – 80a4b + 80a3b2 – 40a2b3 +
32a
2. (2a –
10ab4 – b5
A binomial experiment has 4 trials, with p = 0.3.
b)5
3. What is the probability of 1 success? 0.4116
4. What is the probability of at least 2 successes?
0.3483
Holt Algebra 2
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Lesson Quiz: Part II
A binomial experiment has 4 trials, with p = 0.3.
5. There is a 10% chance that Nila will have to wait for
a train to pass as she heads for school. What is the
probability that she will not have to wait for a train
about 59%
all 5 days this week?
6. Krissy has 3 arrows. The probability of her hitting
the target is . What is the probability that she will
get at least one arrow on the target?
78.4%
Holt Algebra 2