Binomial Distributions
• How do we use the Binomial Theorem to
expand a binomial raised to a power?
•How do we find binomial probabilities
and test hypotheses?
Holt McDougal Algebra 2
Binomial Distributions
Problem-Solving Application
4 Steps for Problem Solving
1
Understand the Problem
2
Make a Plan
3
Solve
4
Look Back
Holt McDougal Algebra 2
Binomial Distributions
Example 1: 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?
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
= 0.8.
Binomial Distributions
Example 1: 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?
2
Make a Plan
The direct way to solve the problem is to
calculate P(3) + P(4).
Holt McDougal Algebra 2
Binomial Distributions
Example 1: 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?
3
Solve
P(3) + P(4)
= 4C3 (0.80)3 (0.20)1 + 4C4 (0.80)4 (0.20)0
 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 McDougal Algebra 2
Binomial Distributions
Example 1: 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?
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 McDougal Algebra 2
Binomial Distributions
Example 2: Problem-Solving Application
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?
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
Example 2: Problem-Solving Application
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?
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."
Holt McDougal Algebra 2
Binomial Distributions
Example 2: Problem-Solving Application
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?
3
Solve
Step 1 Find P(0 or 1 correct).
P(0) + P(1)
= 20C0 (0.25)0 (0.75)20 + 20C1 (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 McDougal Algebra 2
Binomial Distributions
Example 2: Problem-Solving Application
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?
4
Look Back
The answer is reasonable since it is less than
but close to 1.
Holt McDougal Algebra 2
Binomial Distributions
Example 3: Problem-Solving Application
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?
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
Holt McDougal Algebra 2
Binomial Distributions
Example 3: Problem-Solving Application
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?
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.
Holt McDougal Algebra 2
Binomial Distributions
Example 3: Problem-Solving Application
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?
3
Solve
Step 1 Find P(24 or 25 acceptable parts).
P(24) + P(25)
= 25C24 (0.98)24 (0.02)1 + 25C25 (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.
Holt McDougal Algebra 2
Binomial Distributions
Example 3: Problem-Solving Application
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?
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 McDougal Algebra 2
Binomial Distributions
Lesson 3.3 Practice C
Holt McDougal Algebra 2