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

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Senior Manager
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A and B alternately toss a coin. The first one to turn up a [#permalink]  05 Jul 2004, 15:43
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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.
Senior Manager
Joined: 21 Mar 2004
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Location: Cary,NC
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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 ?
_________________

ash
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I'm crossing the bridge.........

Senior Manager
Joined: 21 Mar 2004
Posts: 444
Location: Cary,NC
Followers: 2

Kudos [?]: 20 [0], given: 0

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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ash
________________________
I'm crossing the bridge.........

GMAT Club Legend
Joined: 15 Dec 2003
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Is it:
1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9
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Best Regards,

Paul

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 26

Kudos [?]: 216 [0], given: 0

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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Best Regards,

Paul

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