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A and B alternately toss a coin. The first one to turn up a

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boksana
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A and B alternately toss a coin. The first one to turn up a [#permalink] New post   05 Jul 2004, 16:43
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A and B alternately toss a coin. The first one to turn up a head wins. If no more than five tosses each are allowed for a single game find the probability that the person who tosses first will win the game.



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Re: A and B alternately toss a coin. The first one to turn up a [#permalink] New post   05 Jul 2004, 16:55
The first person has 3 chances to play.

total outcoms = 2^3 = 8

favourable outcomes = _ _ H = 2^2 = 4

probability of first guy winning is 4/8 = 1/2

right ?
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Re: A and B alternately toss a coin. The first one to turn up a [#permalink] New post   05 Jul 2004, 17:38
ashkg wrote:
The first person has 3 chances to play.

total outcoms = 2^3 = 8

favourable outcomes = _ _ H = 2^2 = 4

probability of first guy winning is 4/8 = 1/2

right ?


I hope this time I am right.

probability of one guy getting the first chance = 1/2
probability of this guy getting a heads in the last throw = 1/2

final probability = 1/2 * 1/2 = 1/4

Boksana,
if this is wrong pls post the solution :)

- ash
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Re: A and B alternately toss a coin. The first one to turn up a [#permalink] New post   05 Jul 2004, 18:54
Is it:
1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9
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Re: A and B alternately toss a coin. The first one to turn up a [#permalink] New post   05 Jul 2004, 19:22
We calculate probability according to first person's throws
1st throw: probability of getting H and win is 1/2
2nd throw: probability of A getting T, B getting T and A getting H is (1/2)^3
3rd throw: probability of A getting T, B getting T, A getting T again, B getting T again, then A finally getting H on third throw is: (1/2)^5
4th throw: same process and you get (1/2)^7
5th throw: same process and you get (1/2)^9

Probability of first thrower winning is just sum of individual probabilities.



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Re: A and B alternately toss a coin. The first one to turn up a [#permalink]
05 Jul 2004, 19:22
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