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Unit No.5.
PROBABILITY
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
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302-DECISION SCIENCE
Unit No.5. Probability
5.2 Case 2: Probability
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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Probability
The Sample Space in this case would be-
{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
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Probability
• Total Number of possible outcomes, n= 36
i) Let A- Event that the sum is more than 9 i.e. the
sum is 10 or 11 or 12
The favorable outcomes in this case are-
(4,6) (5,5) (5,6) (6 4) (6 5) (6 6)
Hence Probability is-
• 𝑷 𝑩 = 𝒑 =
𝒎
𝒏
=
𝟔
𝟑𝟔
=
𝟏
𝟔
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Probability
• Total Number of possible outcomes, n= 36
ii) Let B- Event that the sum is multiple of 3
The favorable outcomes in this case are-
(1,2) (1,5) (2,1) (2,4) (3,3) (3,6) (4,2) (4,5) (5,1) (5,4)
(6,3) (6,6) i.e. m=12
Hence Probability is-
• 𝑷 𝑩 = 𝒑 =
𝒎
𝒏
=
𝟏𝟐
𝟑𝟔
=
𝟏
𝟑
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Probability
iii) Let C- Event that the sum is multiple of 4
The favorable outcomes in this case are-
(1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6,6) i.e. m=9
Hence Probability is-
• 𝑷 𝑪 = 𝒑 =
𝒎
𝒏
=
𝟗
𝟑𝟔
=
𝟏
𝟒
Now Probability of Sum is divisible by both 3 & 4 is
𝑷 𝑩Ո𝑪 = 𝒑 =
𝒎
𝒏
=
𝟏
𝟑𝟔
Now the Probability that the sum is divisible by 3 or 4 is
given as-
𝑃 𝐵Ս𝐶 = 𝑃 𝐵 + 𝑃 𝐶 − 𝑃 𝐵Ո𝐶 =
1
3
+
1
4
−
1
36
=
𝟓
𝟗