A and B toss a coin alternately till one of them gets a head and wins the game. if A starts first , find the probability that B will win the game.

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U.k. Akshay posted a question in Class XI-XII Tuition (PUC) Question: A and B toss a coin alternately till one of them gets a head and wins the game. if A starts first , find the probability that B will win the game. Answer: If ‘H’ denotes the event of head and ‘T’ denotes the event of tail, then the...
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U.k. Akshay posted a question in Class XI-XII Tuition (PUC) Question: A and B toss a coin alternately till one of them gets a head and wins the game. if A starts first , find the probability that B will win the game. Answer: 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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if B win the game then then probability would be : (0.5)(0.5) + (0.5)^4 + (0.5)^6 +.... and so on. which is equal to 1/3. in the series, the first 0.5 is for loosing the game by A and second 0.5 is for winning the game by B and so on.
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A player has a 2/3 chance of winning, B's chances of winning are therefore 1/3
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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.
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