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
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 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
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 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?
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%.
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?
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
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
Example 3 Continued
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 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 McDougal Algebra 2
Binomial Distributions
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 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
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."
Holt McDougal Algebra 2
Binomial Distributions
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 McDougal Algebra 2
Binomial Distributions
Check It Out! Example 3a Continued
4
Look Back
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
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
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.
Holt McDougal Algebra 2
Binomial Distributions
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.
Holt McDougal Algebra 2
Binomial Distributions
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 McDougal Algebra 2