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Exercise 12.5 · Q19

Q.There are three events AA, BB and CC, of which one and only one can happen. If the odds are 7 to 4 against AA and 5 to 3 against BB, then the odds against CC are

(1) 23:6523:65
(2) 65:2365:23
(3) 23:8823:88
(4) 88:2388:23
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Step 1. Convert odds to probabilities. Odds 7 to 4 AGAINST AA means P(A)=47+4=411P(A)=\dfrac4{7+4}=\dfrac4{11}. Odds 5 to 3 against BB means P(B)=35+3=38P(B)=\dfrac3{5+3}=\dfrac38.

Step 2. Use exhaustiveness. Since exactly one of A,B,CA,B,C happens, P(A)+P(B)+P(C)=1P(A)+P(B)+P(C)=1. Using LCD 88: P(A)=411=3288P(A)=\dfrac4{11}=\dfrac{32}{88}, P(B)=38=3388P(B)=\dfrac38=\dfrac{33}{88}, so P(C)=1−3288−3388=88−32−3388=2388P(C)=1-\dfrac{32}{88}-\dfrac{33}{88}=\dfrac{88-32-33}{88}=\dfrac{23}{88}. …

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