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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > Two persons a and B toss an unbiased coin alternatively...
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Two persons a and B toss an unbiased coin alternatively on the understanding that the first who gets the head wins.If A starts the game, find their respective chances of winning.

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'A' can win in odd trials. Let the probability of A getting heads be 'p' So '1-p' is the probability that 'A' loses. If 'A' wins in 1st trail then it is simply = p for 'A' to win in 3rd it has to lose in 1st two and win in 3rd = (1-p)^2 *p..and so on. So prob of A winning = p + (1-p)^2*p + (1-p)^4*p+... ...
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'A' can win in odd trials. Let the probability of A getting heads be 'p' So '1-p' is the probability that 'A' loses. If 'A' wins in 1st trail then it is simply = p for 'A' to win in 3rd it has to lose in 1st two and win in 3rd = (1-p)^2 *p..and so on. So prob of A winning = p + (1-p)^2*p + (1-p)^4*p+... = p [1+(1-p)^2 + (1-p)^4+... ] p=1/2 So, probability of 'A winning' = 1/2 * [4/(4-1)] [summing the geometric series] = 2/3 Thus, probability of 'B' winning is 1/3 read less
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

A: 2/3 and B:1/3
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Basic knowledge is a pillar of success

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If ‘H’ denotes the event of head and ‘T’ denotes the event of tail, then the sample space for B getting a head is an infinite series: T H T T T H T T T T T H T T T T T T T H … As the probability of obtaining T and H are identically (1/2) at a tossing of the coin, the overall probability that...
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If ‘H’ denotes the event of head and ‘T’ denotes the event of tail, then the sample space for B getting a head is an infinite series: T H [T (when A tossed the coin) followed by H (when B tossed the coin)] T T T H [T (when A tossed the coin) followed by T (when B tossed the coin) T (when A tossed the coin for second time) followed by H (when B tossed the coin for a second time)] T T T T T H T T T T T T T H … As the probability of obtaining T and H are identically (1/2) at a tossing of the coin, the overall probability that B will obtain head first is P = [(1/2)*(1/2)] + [(1/2)*(1/2)* (1/2)*(1/2)] + [(1/2)*(1/2)* (1/2)*(1/2)* (1/2)*(1/2)*] + … = (1/2)^2 + (1/2)^4+(1/2)^6 + … = (1/4) + (1/4)^2+(1/4)^3 + … (sum of successive, infinite terms of a geometric progression with first member =1/4 and common factor also =1/4) Therefore P = (1/4)*[1/(1-1/4)]= (1/3) read less
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2/3 and 1/3.
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