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By Fazlollah M. Reza

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The number of favorable cases is Therefore the probability in question is 39) . (52) (2 . 2 = 39! 50! 37! 52! = 39 ·38 = 19 51 . 52 34 2-14. Trees and State Diagrams. The material of this section is intended to offer a graphical interpretation for certain simple problems of probability which arise in dealing with repeated trials of an experiment.

Solution. This is an example of a situation where Bayes's theorem can be applied. Let E be the event that a black ball has been drawn; Ai is the event that the ith urn has been chosen, i = 1, 2, 3. Then P{EIAd = >~ P{EIA 2 l = ;~ P{EIA s} = >~ P{AaIE} = P{choosing urn Uslblack ball drawn} Also, P{As}P{EIA a} 3 1 P{EIA,}PfAd i=l }~(~'3 ~~ . ~'2 + 7& + ~~) 15 = 37 Example 2-21. Three urns are given: Urn 1 contains two white, three black, and four red balls. Urn 2 contains three white, two black, and two red balls.

Solution. The main assumption in this and in similar problems is the concept of independence of successive trials and the equally probable outcomes. Let A and B be the events of getting no tail and exactly one tail, respectively. Then PIA} = PIB} nr = 10 The events of interest are or 1,~24 = 1,~~4 U -- A = A' (a) PIA'} (b) = U - (A PIA' 1 1 1,023 = - 1,024 = 1,024 + B) = (U - A) - B A' _ B} = 1,023 _ ~ = 1,013 =; 1,024 1,024 B 1,024 2-10. Conditional Probability. Consider two events A and B. The conditional probability of event A based on the hypothesis that event B has occurred is defined by the following relation: P{AIB} = P{AB} P{B} P{B} ¢ 0 The use of this definition can be justified by returning to the previously treated case of Sec.

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