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EMIS 7300 Team Assignment 6 Solutions. Problem 2-20 Random Events –J1 = junior on draw 1 –J2 = junior on draw 1 –S1 = sophomore on draw 2 –S2 = sophomore.

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Presentation on theme: "EMIS 7300 Team Assignment 6 Solutions. Problem 2-20 Random Events –J1 = junior on draw 1 –J2 = junior on draw 1 –S1 = sophomore on draw 2 –S2 = sophomore."— Presentation transcript:

1 EMIS 7300 Team Assignment 6 Solutions

2 Problem 2-20 Random Events –J1 = junior on draw 1 –J2 = junior on draw 1 –S1 = sophomore on draw 2 –S2 = sophomore on draw 2 –Note that draw 2 is not independent from draw 2 a: P(J1) = 3/10 = 0.3 b: P(J2|S1) = 3/10 = 0.3 c: P(J2|J1) = 8/10 = 0.8 d: P(S1 and S2) = P(S1)P(S2|S1) = (0.7)(0.7) = 0.49 e: P(J1 and J2) = P(J1)P(J2|J1) = (0.3)(0.8) = 0.24

3 Problem 2-20 Continued f: P(S1 and J2) + P(J1 and S2) = P(S1)P(J2|S1) + P(J1)P(S2|J1) = (0.7)(0.3) + (0.3)(0.2) = 0.27.

4 Problem 2-21 Random Events –A = solider is at Abu Ilan –E = solider is at El Kamin –B = sees Bedouin –F = sees Farima

5 Problem 2-21: Probability Tree 0.5 A F|A E P(A and B) = 0.25 P(A and F) = 0.25 B|A B|E F|E P(E and B) = 0.4 P(E and F) =

6 Problem 2-21 Continued P(B) = P(A and B) + P(E and B) = = 0.65 P(A|B) = P(B|A)P(A)/P(B) = 0.5*0.5/0.65 = P(E|B) = 1 – P(A|B) = 1 – = 0.615

7 Problem 2-22 P(BB|A) = 0.5*0.5 = 0.25 P(BB|E) = 0.8*0.8 = 0.64 P(A and BB) = 0.5 * 0.25 = P(E and BB) = 0.5 * 0.64 = P(BB) = = P(A|BB) = P(A and BB)/P(BB) = P(E|BB) = 1 – = 0.719

8 Problem 2-23 P(adjusted) = 0.8 (prior probability) P(pass|adjusted) = 0.9 P(pass|not adjusted) = 0.2 P(adjusted|pass) = ?

9 Problem 2-23: Probability Tree Pass 0.2 Not Adjusted P(Adjusted and Pass) = 0.72 P(Adjusted and Not Pass) = Not pass Pass Not Pass Adjusted P(Not Adjusted and Pass) = 0.04 P(Not Adjusted and Not Pass) = 0.16

10 Problem 2-23 Continued P(Pass) = = P(Adjusted|Pass) = P(Adjusted and Pass)/P(Pass) =0.72/0.76 = 0.95


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