Question

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 α,β and γ be, respectively, the probabilities that A,E and C gets the first head. Prove that β=(1-p)α Determine α,β and γ (in terms of p). Select the correct answer from above options

Answer 1

Correct Answer - A::B::C Let q=1-p= probability of getting the tail. We have, α= probability of A getting the head on tossing firstly =P(H1orT1T2T3H4orT1T2T3T4T5T6H7or...) =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))