Binomial lecture

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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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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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• 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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-( ) ( ) x n x

n xP X x P x C p q

Determining x successes in n trials:

where, = number of trails = probability of a success = probability of a failure = number of successes

!

!( - )!n x

npqx

nC

x n x

Calculating the Binomial Probability


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• 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

Are the conditions required for the binomial experiment met?

• 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

The Binomial Experiment - Example

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• Let X be the binomial random variable indicating

the number of late students

0 5-0

5 0P(X = 0) = P(0) = C (0.10) (0.90) = 0.5905

5 5-5

5 5P(X = 5) = P(5) = C (0.10) (0.90) = 0.00001


Calculate the probability that zero students are late Calculate the probability that one student is late Calculate the probability that three students are late

1 5-1

5 1P(X = 1) = P(1) = C (0.10) (0.90) = 0.3281

3 5-3

5 3P(X = 3) = P(3) = C (0.10) (0.90) = 0.008

2 5-2

5 2P(X = 2) = P(2) = C (0.10) (0.90) = 0.072

4 5-4

5 4P(X = 4) = P(4) = C (0.10) (0.90) = 0.00045

-( ) ( ) (1 - )x n x

n xP X x p x C p p

• Let X be the binomial random variable indicating

the number of late students


0 5-0

5 0P(X = 0) = P(0) = C (0.10) (0.90) = 0.5905

5 5-5

5 5P(X = 5) = P(5) = C (0.10) (0.90) = 0.00001

1 5-1

5 1P(X = 1) = P(1) = C (0.10) (0.90) = 0.3281

3 5-3

5 3P(X = 3) = P(3) = C (0.10) (0.90) = 0.008

2 5-2

5 2P(X = 2) = P(2) = C (0.10) (0.90) = 0.072

4 5-4

5 4P(X = 4) = P(4) = C (0.10) (0.90) = 0.00045

X P(X)

0 0.5905

1 0.3281

2 0.0729

3 0.0081

4 0.00045

5 0.00001

∑P(X) ≈ 1

• Calculate the probability that 2 or less students

will be late


X P(X)

0 0.5905

1 0.3281

2 0.0729

3 0.0081

4 0.00045

5 0.00001

∑P(X) ≈ 1

P(X ≤ 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= 0.5905 + 0.3281 + 0.0729

= 0.9915


X P(X)

0 0.5905

1 0.3281

2 0.0729

3 0.0081

4 0.00045

5 0.00001

∑P(X) ≈ 1

P(X < 2)

= P(X = 0) + P(X = 1)

= 0.5905 + 0.3281

= 0.9186

• Calculate the probability that less than 2 students

will be late


X P(X)

0 0.5905

1 0.3281

2 0.0729

3 0.0081

4 0.00045

5 0.00001

∑P(X) ≈ 1

P(X ≥ 4)

= P(X = 4) + P(X = 5)

= 0.00045 + 0.00001

= 0.00046

• Calculate the probability that 4 or more than 4

students will be late


X P(X)

0 0.5905

1 0.3281

2 0.0729

3 0.0081

4 0.00045

5 0.00001

∑P(X) ≈ 1

P(X > 4)

= P(X = 5)

= 0.00001

• Calculate the probability that more than 4

students will be late


X P(X)

0 0.5905

1 0.3281

2 0.0729

3 0.0081

4 0.00045

5 0.00001

∑P(X) ≈ 1

P(X ≥ 3)

= P(X = 3) + P(X = 4) + P(X = 5)

= 0.00856 OR

= 1 – P(X ≤ 2)

= 1 – 0.9915

= 0.0085

• Calculate the probability that 3 or more students

will be late


X P(X)

0 0.5905

1 0.3281

2 0.0729

3 0.0081

4 0.00045

5 0.00001

∑P(X) ≈ 1

P(X > 3)

= P(X = 4) + P(X = 5)

= 0.00046 OR

= 1 – P(X ≤ 3)

= 1 – 0.9996

= 0.0004

• Calculate the probability that more than 3 students

will be late

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• 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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– What is the expected number of students that come

late?

– What is the standard deviation for the number of

students who come late?

( ) 5(0.10) 0.5E X np

2 ( ) 5(0.10)(0.90) 0.67Var X npq


Binomial lecture

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