The pattern in the table can help you expand any binomial by using the Binomial Theorem .
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Holt McDougal Algebra 2 Binomial Distributions The pattern in the table can help you expand any binomial by using the Binomial Theorem .
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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. - PowerPoint PPT Presentation
Transcript of The pattern in the table can help you expand any binomial by using the Binomial Theorem .
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 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
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) = 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 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 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?
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
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 McDougal Algebra 2
Binomial Distributions
22 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
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 McDougal Algebra 2
Binomial 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 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
11 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
22 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
Solve33
= 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 probability that Wendy will get at least 2 answers correct is about 0.98.
Holt McDougal Algebra 2
Binomial Distributions
Check It Out! Example 3a Continued
Look Back44
The answer is reasonable since it is less than but close to 1.
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?
Holt McDougal Algebra 2
Binomial Distributions
Check It Out! Example 3b Continued
11 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
Holt McDougal Algebra 2
Binomial Distributions
22 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.
Holt McDougal Algebra 2
Binomial Distributions
Solve33
= 25C24(0.98)24(0.02)25-24 + 25C25(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
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.
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Check It Out! Example 3b Continued
