The document describes the binomial experiment, which involves repeating a trial with two possible outcomes (success or failure) a fixed number of times. Each trial has a known probability of success (p) or failure (q=1-p) and trials are independent. The number of successes is then modeled by the binomial distribution. An example calculates the probabilities of different numbers of students being late to class when sampling 5 students, where the probability of an individual student being late is 10%. Key aspects of the binomial experiment and calculating binomial probabilities are summarized.

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2. The Binomial Experiment
• Repeated n times(trials) under identical
conditions
• Each trial can result in only one out of two
outcomes
– Success – probability success p
– Failure – probability failure q = 1 – p
• Trials are independent
• Measure number of successes, x, in n trails
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3. The Binomial Experiment
Typical cases where the binomial experiment
applies:
– A coin flipped results in heads or tails
– A party wins or loses election
– An employee is male or female
– A car uses leaded, or unleaded fuel
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4. The Binomial Experiment
• Binomial distribution is the probability distribution
that applies to the binomial experiment
• Displayed in the form of a table where the first
row (or column) displays all possible number of
successes, second row (or column) displays the
probability associated with number of successes
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5. The Binomial Experiment
Calculating the Binomial Probability
Determining x successes in n trials:
P ( X  x)  P ( x)  n
x
Cx p q n-x
where, n = number of trails
p = probability of a success
q = probability of a failure
x = number of successes
n!
n Cx 
x !(n - x )! 5

6. The Binomial Experiment - Example
• 10% of students are late for the early morning class
• In a sample of 5 students, find the probability
distribution of the number students that are late
Are the conditions required for the binomial experiment met?
• Repeated n = 5 times
• Each trial can result in only one out of two outcomes
– Success – late for class → p = 0.10
– Failure – not late for class → q = 1 - 0.10 = 0.90
• Students are independent 6

7. The Binomial Experiment - Example
• Let X be the binomial random variable indicating
the number of late students
Calculate the probability that threestudents are late
Calculate the probability that zerostudent is late late
one students are
P ( X  x)  p( x)  n C x p x (1 - p ) n - x
P(X = 0) = P(0) = 5 C 0 (0.10) 0 (0.90) 5-0 = 0.5905
P(X = 1) = P(1) = 5 C 1 (0.10) 1 (0.90) 5-1 = 0.3281
P(X = 2) = P(2) = 5 C 2 (0.10) 2 (0.90) 5-2 = 0.072
P(X = 3) = P(3) = 5 C 3 (0.10) 3 (0.90) 5-3 = 0.008
P(X = 4) = P(4) = 5 C 4 (0.10) 4 (0.90) 5-4 = 0.00045
P(X = 5) = P(5) = 5 C 5 (0.10) 5 (0.90) 5-5 = 0.00001 7

8. The Binomial Experiment - Example
• Let X be the binomial random variable indicating
the number of late students
X P(X)
P(X = 0) = P(0) = 5 C 0 (0.10) 0 (0.90) 5-0 = 0.5905 0 0.5905
P(X = 1) = P(1) = 5 C 1 (0.10) 1 (0.90) 5-1 = 0.3281 1 0.3281
P(X = 2) = P(2) = 5 C 2 (0.10) 2 (0.90) 5-2 = 0.072 2 0.0729
P(X = 3) = P(3) = 5 C 3 (0.10) 3 (0.90) 5-3 = 0.008 3 0.0081
P(X = 4) = P(4) = 5 C 4 (0.10) 4 (0.90) 5-4 = 0.00045 4 0.00045
P(X = 5) = P(5) = 5 C 5 (0.10) 5 (0.90) 5-5 = 0.00001 5 0.00001
∑P(X) ≈ 1

9. The Binomial Experiment - Example
• Calculate the probability that 2 or less students
will be late
X P(X)
P(X ≤ 2) 0 0.5905
= P(X = 0) + P(X = 1) + P(X = 2) 1 0.3281
2 0.0729
= 0.5905 + 0.3281 + 0.0729
3 0.0081
= 0.9915 4 0.00045
5 0.00001
∑P(X) ≈ 1

10. The Binomial Experiment - Example
• Calculate the probability that less than 2 students
will be late
X P(X)
P(X < 2) 0 0.5905
= P(X = 0) + P(X = 1) 1 0.3281
2 0.0729
= 0.5905 + 0.3281
3 0.0081
= 0.9186 4 0.00045
5 0.00001
∑P(X) ≈ 1

11. The Binomial Experiment - Example
• Calculate the probability that 4 or more than 4
students will be late
X P(X)
P(X ≥ 4) 0 0.5905
= P(X = 4) + P(X = 5) 1 0.3281
2 0.0729
= 0.00045 + 0.00001
3 0.0081
= 0.00046 4 0.00045
5 0.00001
∑P(X) ≈ 1

12. The Binomial Experiment - Example
• Calculate the probability that more than 4
students will be late
X P(X)
P(X > 4) 0 0.5905
= P(X = 5) 1 0.3281
2 0.0729
= 0.00001
3 0.0081
4 0.00045
5 0.00001
∑P(X) ≈ 1

13. The Binomial Experiment - Example
• Calculate the probability that 3 or more students
will be late
X P(X)
P(X ≥ 3) 0 0.5905
= P(X = 3) + P(X = 4) + P(X = 5) 1 0.3281
= 0.00856 OR 2 0.0729
3 0.0081
= 1 – P(X ≤ 2)
4 0.00045
= 1 – 0.9915 5 0.00001
= 0.0085 ∑P(X) ≈ 1

14. The Binomial Experiment - Example
• Calculate the probability that more than 3 students
will be late
X P(X)
P(X > 3) 0 0.5905
= P(X = 4) + P(X = 5) 1 0.3281
= 0.00046 OR 2 0.0729
3 0.0081
= 1 – P(X ≤ 3)
4 0.00045
= 1 – 0.9996 5 0.00001
= 0.0004 ∑P(X) ≈ 1

15. The Binomial Experiment
• Mean and standard deviation of binomial
random variable
  E( X )  np    2  Var ( X )  npq
REMEMBER
Repeated n times(trials) under identical conditions
Each trial can result in only one out of two outcomes
Success – probability success p
Failure – probability failure q = 1 – p
Measure number of successes, x, in n trails
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16. The Binomial Experiment - Example
– What is the expected number of students that come
late?
  E( X )  np  5(0.10)  0.5
– What is the standard deviation for the number of
students who come late?
    Var ( X )  npq  5(0.10)(0.90)  0.67
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