Binomial Lecture

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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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