InterviewSolution
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InterviewSolution
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Three players A,B and C toss a coin cyclically in that order (i.e. A,B,C,A,B,C,A,B,……) till a head shows. Let p be the probability that the coin shows a head. Let `alpha,beta` and `gamma` be, respectively, the probabilities that A,E and C gets the first head. Prove that `beta=(1-p)alpha` Determine `alpha,beta` and `gamma` (in terms of p). |
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Answer» Correct Answer - A::B::C Let `q=1-p=` probability of getting the tail. We have, `alpha=` probability of A getting the head on tossing firstly `=P(H_(1)orT_(1)T_(2)T_(3)H_(4)or T_(1)T_(2)T_(3)T_(4)T_(5)T_(6)H_(7)or...)` `=P(H)+P(H)P(T)^(3)+P(H)P(T)^(6)+...` `=(P(H))/(1-P(T)^(3))=(P)/(1-q^(3))` Also, `beta=` probability of B getting the head on tossing secondly `=P(T_(1)H_(2)orT_(1)T_(2)T_(3)T_(4)T_(5)orT_(1)T_(2)T_(3)T_(4)T_(5)T_(6)T_(7)T_(8)or...)` `=P(H)[P(T)+P(H)P(T)^(4)+P(H)P(T)^(7)+...]` `=P(T)[P(H)+P(H)P(T)^(3)+P(H)P(T)^(6)+...]` `=qalpha=(1-p)alpha=(p(1-p))/(1-q^(3))` Again, we have `alpha+beta+gamma=1` `rArr " "gamma=1-(alpha+beta)=1-(p+p(1-p))/(1-q^(2))` `-1-(p+p(1-p))/(1-(1-p)^(3))` `=(1-(1-p)^(3)-p-p(1-p))/(1-(1-p)^(2))` `gamma=(1-(1-p)^(3)-2p+p^(2))/(1-(1-p)^(3))=(p-2p^(2)+p^(2))/(1-(1-p)^(3))` Also, `alpha=(p)/(1-(1-p)^(3)),beta=(p(1-p))/(1-(1-p)^(3))` |
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