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Consider we toss a coin repeatedly.The coin is unfair and p(H) =p.The game ends the first time 2 consecutive heads(HH) or 2 consecutive tails(TT) are observed.We win if HH is observed and loose if TT is observed.For eg if outcome is HTHTT we loose, on the other hand if outcome is THTHTHH we win.prob that we win?

a)p2(1+q)/(1-pq)

b)p2(1-q)/(1-pq)

c)p2(1+q)/(1-p)

d)p2(1+q)/1+pq

1 Answer

Best answer
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Here there are 2 cases..Let the probability of head be 'p' as given in question and that of tail be 'q'..Lets find probability for each of the cases and then we sum it up :

Case 1 : The toss begins with head :

In that case it is possible that first 2 head comes and the game is ended..Else one head , one tail then 2 heads n then game is won and so on..

So P(win in 1st case)    =   p2  [First two tosses lead to head] +  pqp2 [First head , then tail , then two heads in a row]                                                         +   pqpqp2   + ...

                                   =   p2  [ 1 + pq + (pq)2 + .......... ]

                                   =   p2  /  (1 - pq)    [As the internal terms form an infinite G.P with initial term = 1 and common ratio = pq]

Case 2 :  Toss begins with tail :

So here it is possible that first is tail then two heads and hence game  ends and results in win..Else a tail , then a head then a tail then two heads in a row and hence game ends here and so on..

Hence P(win in 2nd case)    =  qp2  + qpqp2 + ..........

                                         =  qp2 [ 1 + qp + (qp)2 + .......]

                                         =  qp2  /  (1 - pq)

Hence P(win)                     = P(win in case 1)  + P(win in case 2)

                                        =  [p2  /  (1 - pq)]  +  [qp2  /  (1 - pq)]

                                        =  p(1 + q) / (1 - pq)

Hence A) should be the correct answer..

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