Section 2.4:Conditional Probability1DefinitionFor any two events A and B with P (B) > 0, theconditional probability of A given that B has oc-cured isP (A|B) =P (A ∩ B)P (B).2Rules for Conditional Probability• Multiplication RuleP (A ∩ B) = P (A|B)P (B).• The law of total probability: assume A1, · · · , Akare mutually exclusive and exhaustive events.That is, there is no paired overlap and theirunion is S, orAi∩ Aj= φ, i 6= jandA1∪ A2∪ · · · ∪ Ak= S.ThenP (B) =P (B|A1)P (A1) + P (B|A2)P (A2)+ · · · + P (B|Ak)P (Ak)=kXi=1P (B|Ai)P (Ai).A special case:P (B) = P (B|A)P (A) + P (B|A0)P (A0).3• Bayes’s Theorem: assume A1, · · · , Akare mu-tually exclusive and exhaustive events. Forany B with P (B) > 0,P (Aj|B) =P (B|Aj)P (Aj)Pki=1P (B|Ai)P (Ai)for j = 1, · · · , k.A special case:P (A|B) =P (B|A)P (A)P (B|A)P (A) + P (B|A0)P (A0).4First example of Section 2.4: 2.25 on textbook.• Digital camera, 60% include an optional mem-ory card, 40% include an extra battery, and30% include both.• Then, P (A) = 0.6, P (B) = 0.40 and P (A ∩B) = 0.30.• Thus,P (A|B) =P (A ∩ B)P (B)=0.30.4= 0.75.andP (B|A) =P (A ∩ B)P (A)=0.30.6= 0.5.5Second example of Section 2.4: 2.26 on text-book.• Suppose P (A) = 0.14, P (B) = 0.23, P (C) =0.37, P (A∩B) = 0.08, P (A∩C) = 0.09, P (B∩C) = 0.13 and P (A ∩ B ∩ C) = 0.05.• The, we haveP (A|B) =P (A ∩ B)P (B)=0.080.23= 0.348.P (A|B ∪ C)=P (A ∩ (B ∪ C))P(B∪C)=P (A ∩ B) + P (A ∩ C) − P (A ∩ B ∩ C)P (B) + P (C) − P (B ∩ C)=0.08 + 0.09 − 0.050.23 + 0.37 − 0.13=0.2553.6P (A ∪ B ∪ C)=P (A) + P (B) + P (C) − P (A ∩ B)− P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C)=0.14 + 0.23 + 0.37 − 0.08 − 0.09 − 0.13 + 0.05=0.49.P (A|at least one)=P (A|A ∪ B ∪ C)=P (A ∩ (A ∪ B ∪ C))P (A ∪ B ∪ C)=P (A)P (A ∪ B ∪ C)=0.140.49=0.2857.andP (A ∪ B|C)=P ((A ∪ B) ∩ C)P (C)=P (A ∩ C) + P (B ∩ C) − P (A ∩ B ∩ C)P (C)=0.09 + 0.13 − 0.050.37=0.4595.7Third example of Section 2.4: 2.29 on textbook.• Let A1= {brand 1}, A2= {brand 2}, A3={brand 3}, and B = {warranty}.• P (A1) = 0.5, P (A2) = 0.3 and P (A3) = 0.2.• P (B|A1) = 0.25, P (B|A2) = 0.2, and P (B|A3) =0.1.• Then,P (A1∩ B) = P (B|A1)P (A1) = 0.125.P (A2∩ B) = P (B|A2)P (A2) = 0.060andP (A3∩ B) = P (B|A3)P (A3) = 0.02.•P (B) = 0.125 + 0.060 + 0.02 = 0.205.• Thus, we haveP (A1|B) =P (A1∩ B)P (B)=0.1250.205= 0.6098P (A2|B) =P (A2∩ B)P (B)=0.060.205= 0.2927P (A3|B) =P (A3∩ B)P (B)=0.020.205= 0.0976.8Fourth example of Section 2.4: 2.31 on textbook.• LetA = {has the disease}andB = {test is positive}.• Suppose P (A) = 0.001, P (B|A) = 0.99, andP (B|A0) = 0.02.• Then, we haveP (A|B)=P (B|A)P (A)P (B|A)P (A) + P (B|A0)P (A0)=0.99 × 0.0010.99 × 0.001 + 0.02 × 0.999=0.0472.• Q: What does this imply?9Fifth example of Section 2.4. Let A, B, C beevents and S be sample space. Suppose P (A) =0.4, P (B) = 0.3, P (C) = 0.2, P (A ∩ B) = 0.11,P (A∩C) = 0.05, P (B ∩C) = 0.06, P (A∩B ∩ C) =0.01.(a) Compute P (A0∪B0), P (B0∪C0), P (A0∪C0) andP (A0∪ B0∪ C0).P (A0∪ B0) = 1 − P (A ∩ B) = 1 − 0.11 = 0.89,P (B0∪ C) = 1 − P (B ∩ C) = 1 − 0.06 = 0.94,P (A0∪ C0) =1 − P (A ∩ C) = 0.95,andP (A0∪ B0∪ C0) =1 − P (A ∩ B ∩ C)=1 − 0.01=0.99.10(b) Compute P (A0∩B0), P (B0∩C0), P (A0∩C0) andP (A0∩ B0∩ C0).First, we haveP (A ∪ B) = 0.4 + 0.3 − 0.11 = 0.59,P (A ∪ C) = 0.4 + 0.2 − 0.05 = 0.55,P (B ∪ C) = 0.3 + 0.2 − 0.06 = 0.54,andP (A ∪ B ∪ C)=0.4 + 0.3 + 0.2 − 0.11 − 0.05 − 0.06 + 0.01=0.69.Thus, we have Note thatP (A0∩ B0) = 1 − P (A ∪ B)1 − 0.59 = 0.41,P (A0∩ C0) = 1 − P (A ∪ C) = 1 − 0.55 = 0.45,P (B0∩ C0) = 1 − P (B ∪ C) = 1 − 0.44 = 0.56,andP (A0∩B0∩C0) = 1−P (A∪B∪C) = 1−0.69 = 0.31.11(c) Compute P (B|A), P (B|A0), P (B0|A), P (B0|A0).P (B|A) =P (A ∩ B)P (A)=0.110.4= 0.275P (B|A0) =P (B ∩ A0)P (A0)=P (B) − P (A ∩ B)1 − P (A)=0.3 − 0.110.6= 0.3167.P (B0|A) =P (B0∩ A)P (A)=P (A) − P (A ∩ B)P (A)=0.4 − 0.110.4= 0.725.P (B0|A0) =P (B0∩ A0)P (A0)=0.410.6= 0.6833.From this problem, we can findP (B|A) + P (B0|A) = 1andP (B|A0) + P (B0|A0) = 1.12(d) Compute P (C|A∩B), P (C|A∩B0), P (C|A0∩B)and P (C|A0∩ B0).P (C|A ∩ B) =P (A ∩ B ∩ C)P (A ∩ B)=0.010.11= 0.0909.P (C|A ∩ B0) =P (A ∩ B0∩ C)P (A ∩ B0)=P (A ∩ C) − P (A ∩ B ∩ C)P (A) − P (A ∩ B)=0.05 − 0.010.4 − 0.11= 0.1379.P (C|A0∩ B) =0.06 − 0.010.3 − 0.11= 0.2632.P (C|A0∩ B0) =P (A0∩ B0∩ C)P (A0∩ B0)=P (A0∩ B0) − P (A0∩ B0∩ C0)P (A0∩ B0)=0.41 − 0.310.41= 0.2439.13(e) Compute P (C0|A ∩ B), P (C0|A ∩ B0), P (C0|A0∩B) and P (C0|A0∩ B0).We use formulaeP (C|A ∩ B) + P (C0|A ∩ B) = 1P (C|A ∩ B0) + P (C0|A ∩ B0) = 1P (C|A0∩ B) + P (C0|A0∩ B) = 1P (C|A0∩ B0) + P (C0|A0∩ B0) = 1Then,P (C0|A ∩ B) = 1 − 0.0909 = 0.9091.P (C0|A ∩ B0) = 1 − 0.1379 = 0.8621.P (C0|A0∩ B) = 1 − 0.2632 = 0.7368.P (C0|A0∩ B0) = 1 − 0.2439 = 0.7541.14Sixth example of Section 2.4.A box has three bags. The first bag has 10 blueballs and 10 red balls. The second bag has 8 blueballs and 2 red balls. The third bag has 12 blueballs and 18 red balls. You are asked to randomlypick up one bag from the box and then randomlypick up one ball from the bag you have just pickedup.Let A1= {first bag}, A2= {second bag}, A3={third bag}, and B = {red ball}.15(a) If the selected bag is the first one, what it theprobability of the color of the ball you see isred.Answer:P (B|A1) =1020= 0.5.(b) If the selected bag is the second one, what itthe probability of the color of the ball you seeis red.Answer:P (B|A2) =210= 0.2.(c) If the selected bag is the third one, what itthe probability of the color of the ball you seeis red.Answer:P (B|A3) =1830= 0.6.16(d) Overall, what is the probability of the color ofthe ball you see is red.Answer: first P (A1) = P (A2) = P (A3) = 1/3.Then, by total law of probability, we haveP (B) …