This document discusses the binomial probability distribution and provides examples of calculating probabilities of coin toss outcomes. Specifically, it addresses: 1) Calculating the probability of getting exactly 5 heads, at least 8 heads, not more than 3 heads, and at least one head when tossing 10 coins. 2) If the 10 coin toss is repeated 50 times, calculating the expected number of times exactly 5 heads will occur. 3) Explaining the case of the binomial probability distribution and providing contact information for the presenter.

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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
5.2.4 Case 1: 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 1: Binomial Probability Distribution

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Probability Distribution
10 Unbiased coins are tossed simultaneously, find the
probability that there will be-
i. Exactly 5 heads
ii. At least 8 heads
iii. Not more than 3 heads
iv. At least one head
v. If this exercise is carried out 50 times, how many times
we can get exactly 5 heads?

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Case 1: Binomial Distribution
Let, p - Probability of getting Head = ½
Similarly, q - Probability of getting Tail = ½
Also the experience is performed using 10 coins i.e. it is
carried out n = 10 number of times.
Hence the probability of getting r successes in 10 trials
is given by Binomial Probability Function as-
P(r) = 10Crprq10-r = 10Cr (
𝟏
𝟐
)r (
𝟏
𝟐
)10-r = 10Cr (
𝟏
𝟐
)10 = 𝟏
𝟏𝟎𝟐𝟒
10Cr

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Case 1: Binomial Distribution
i. Probability of getting 5 heads i.e. r = 5 is-
P(5) =
𝟏
𝟏𝟎𝟐𝟒
10C5
P(5) =
𝟏
𝟏𝟎𝟐𝟒
∗
𝟏𝟎∗𝟗∗𝟖∗𝟕∗𝟔
𝟏∗𝟐∗𝟑∗𝟒∗𝟓
P(5) = 𝟎. 𝟐𝟒𝟔

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Case 1: Binomial Distribution
ii. Probability of getting at least 8 heads i.e. r ≥ 8 is-
P(r ≥ 8) = P r = 8 or 9 or 10 = P 8 + P 9 + P 10
P(r ≥ 8) =
1
1024
(10C8 + 10C9 +10C10)
P(r ≥ 8) =
1
1024
(10C2 + 10C1 +10C0)
P(r ≥ 8) =
1
1024
∗
10∗9
1∗2
+ 10 + 1
P(r ≥ 8) =
1
1024
∗ 56
P(r ≥ 8) = 0.055

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Case 1: Binomial Distribution
iii. Probability of not more than 3 heads i.e. r ≤ 3 is-
P(r ≤ 3) = P r = 0 or 1 or 2 or 3 = P 0 + P 1 + P 2 + P 3
P(r ≤ 3) =
1
1024
(10C0 + 10C1 +10C2 + 10C3)
P(r ≤ 3) =
1
1024
∗ 1 + 10 +
10∗9
1∗2
+
10∗9∗8
1∗2∗3
P(r ≤ 3) =
1
1024
∗ 1 + 10 + 45 + 120 =
176
1024
P(r ≤ 3) = 0.172

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Case 1: Binomial Distribution
iv. At least 1 head i.e. r ≥ 1 is-
P(r ≥ 1) = 1 − P r ˂ 1 = 1 − P 0
P(r ≥ 1) = 1 -
1
1024
(10C0)
P(r ≥ 1) = 1 -
1
1024
P(r ≥ 1) =
1023
1024
P(r ≥ 1) = 0.9

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Case 1: Binomial Distribution
v. Possibility of getting 5 Heads:
If the exercise of throwing n=10 coins is repeated for
N=50 Times,
Then the number of times we expect to get 5 heads
is-
= f(5) = N * P(5)
= 50 * 0.246
=12.3 i.e. 12

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EXERCISE
 Explain case 1: Binomial Probability Distribution

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