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www.sanjivanimba.org.in
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
www.sanjivanimba.org.in
www.sanjivanimba.org.in
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
www.sanjivanimba.org.in
NETWORK ANALYSIS
 At the End of the Session Student will be able to
understand-
A. Case 2: Probability
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Probability
A pair of dice is thrown. Find the probability of
getting the sum-
i) More than 9;
ii) Multiple of 3
iii) Divisible by 3 or 4
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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
www.sanjivanimba.org.in
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
www.sanjivanimba.org.in
EXERCISE
 Solve the Case 2 on Probability.
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For More Details Contact
Dr. V M Tidake
tidkevishal@gmail.com
tidkevishalmba@sanjivani.org.in

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5.5 case 2 probability

  • 1. www.sanjivanimba.org.in 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 www.sanjivanimba.org.in
  • 2. www.sanjivanimba.org.in 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
  • 3. www.sanjivanimba.org.in NETWORK ANALYSIS  At the End of the Session Student will be able to understand- A. Case 2: Probability
  • 4. www.sanjivanimba.org.in Probability A pair of dice is thrown. Find the probability of getting the sum- i) More than 9; ii) Multiple of 3 iii) Divisible by 3 or 4
  • 5. www.sanjivanimba.org.in 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)}
  • 6. www.sanjivanimba.org.in 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
  • 7. www.sanjivanimba.org.in 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
  • 8. www.sanjivanimba.org.in 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
  • 10. www.sanjivanimba.org.in For More Details Contact Dr. V M Tidake tidkevishal@gmail.com tidkevishalmba@sanjivani.org.in