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Q14. A , B and C throw a die alternately till one of them gets a '6' and wins the game. Find their respective probability of wining. If A starts the game.

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Solution

Dear student
Let Ebe the event of getting a six in a single throw of an unbiased die. Then,P(E)=16 and PE¯=1-16=56A wins if he gets a six in 1 st or 4th or 7th ... throw.His probability of getting a 'six' infirst throw=P(E)=16A will get fourth throw if he fails in first,B fails in second and C fails in third throw.So, Probability of winning of A in fourth throw=PE¯E¯E¯E=PE¯PE¯PE¯PE=563×16Similarly, the probability of winning of A in 7th throw=566×16 and so on...Hence probability of winning of A=16+563×16+566×16+...=161-563 Sum of infinte G.P=a1-r and here a=16 and r=563=3691B wins if he gets 'six' in 2nd or 5th or 8th or... throwSo,probability of winning of B=56×16+564×16+567×16=56×161-563 Sum of infinte G.P=a1-r and here a=56×16 and r=563=3091Hence probability of winning of C=1-3691+3091=2591
Regards

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