CASE STUDY ON BINOMIAL PROBABILITY DISTRIBUTION

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Unit No.5.
PROBABILITY DISTRIBUTION
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean, EDP & Associate Professor MBA
1
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in

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302-DECISION SCIENCE
Unit No.5. Probability
5.2.5 Case 2: Binomial
Probability Distribution
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean EDP & Associate Professor MBA
2
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in

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NETWORK ANALYSIS
 At the End of the Session Student will be able to
understand-
A. Case 2: Binomial Probability Distribution

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Probability Distribution
The mean and variance of a binomial distribution are 3 & 2
respectively. Find the probability that the variate takes
values-
i. Exactly 2
ii. At most 2

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Case 2: Binomial Distribution
Given that, np=3 and npq=2 upon dividing both we get,
q = 2/3
Hence,
p = 1 - q
p = 1 - 2/3
P = 1/3
Now, Mean, np = 3 i.e. n=3/p
n = 3/(1/3)
n = 9

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Case 2: Binomial Distribution
i. Probability that variate takes value 2 i.e. r = 2 is-
P(2) = nCrprqn-r = 9C2 (
1
3
)2 (
2
3
)9-2
P(2) =
9∗8
2∗1
(
1
9
) (
2
3
)7
P(2) = 0.235

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Case 2: Binomial Distribution
ii. Probability that variate takes value at max 2 i.e. r ≤ 2
is-
P(r ≤ 2) = P r = 0 or 1 or 2 = P 0 + P 1 + P 2
P(r ≤ 2) = 9C0 (
1
3
)0 (
2
3
)9 + 9C1 (
1
3
)1 (
2
3
)8 + 9C2 (
1
3
)2 (
2
3
)7
P(r ≤ 2) = 0.378

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EXERCISE
 Explain case 2: Binomial Probability Distribution

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For More Details Contact
Dr. V M Tidake
tidkevishal@gmail.com
tidkevishalmba@sanjivani.org.in