Question 712389
 A AND B PLAY 12 GAMES OF CHESS,OF WHICH 
6 ARE WON BY A,
4 BY B AND 
2 END IN A TIE.
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P(A win) = 6/12 = 1/2
P(B win) = 4/12 = 1/3
P(tie) = 2/12 = 1/6
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THEY AGREE TO PLAY 3 MORE GAMES.FIND THE PROBABILITY THAT
a)P(A WIN ALL THE THREE GAMES) = (1/2)^3 = 1/8
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b)P(TWO GAMES END IN A TIE) = (1/6)^2 = 1/36
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c)A AND B WIN ALTERNATIVELY 
P(ABA or BAB) = (1/2)^2(1/3) + (1/3)^2(1/2) = 1/12 + 1/18 = (3+2)/36 = 5/36
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d)P(B WINS AT LEAST ONE GAME) = [P(BB'B')]^3 =[(1/3(1/2)^2]^3 = 1/12^3 
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Cheers,
Stan H.