Binomial Distributions
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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
3
C
2 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 1: 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) = n
C r p r q n-r
n
r
p
q
P ( 1 ) =
3
C
1

3
(0.5) 1 (0.5) 31

0 .
25


0 .
3 7 5
The probability that Jean will make exactly one free throw is 37.5%.
Holt McDougal Algebra 2

Binomial Distributions
Example 2: 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 )


3
C
1
(0.5) 1 (0.5) 31
3

0 .
25
 
3
+
3
C
2
(0.5) 2 (0.5) 32 +
3
C
3
(0.5) 3 (0.5) 33

0 .
25
  

1

0 .
125
  
0 .
3 7 5

0 .
3 7 5

0 .
1 2 5

0 .
8 7 5
The probability that Jean will make at least one free throw is 87.5%.
Holt McDougal Algebra 2

Binomial Distributions
Example 3: Finding Binomial Probabilities
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.

3


2

2
3

2
9

0 .
2 2
The probability that Counselor Jenkins will get 2 of the next 3 students assigned is about 22%.
Holt McDougal Algebra 2

Binomial Distributions
Example 4: Finding Binomial Probabilities
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 )
5
C
2
(0.25) 2 (0.75) 52 +
5
C
3
(0.25) 3 (0.75) 53 +
5
C
4
(0.25) 4 (0.75) 54
+

5
C
5
(0.25) 5 (0.75) 55
0 .
2 6 3 7

0 .
0 8 7 9

0 .
0 1 4 6

0 .
0 0 1 0
Holt McDougal Algebra 2

Binomial Distributions
Lesson 3.3 Practice B
Holt McDougal Algebra 2