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

Oldest

Best Reply

Author

Message

TAGS:

Senior Manager

Joined: 01 May 2004

Posts: 363

Location: USA

Followers: 1

Kudos [?]: 1 [0], given: 0

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

05 Jul 2004, 16:43

00:00

Question Stats:

0% (00:00) correct

0% (00:00) wrong

based on 0 sessions

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

Posts: 447

Location: Cary,NC

Followers: 2

Kudos [?]: 2 [0], given: 0

[#permalink]

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 ?
_________________

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

Senior Manager

Joined: 21 Mar 2004

Posts: 447

Location: Cary,NC

Followers: 2

Kudos [?]: 2 [0], given: 0

[#permalink]

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
_________________

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

GMAT Club Legend

Joined: 15 Dec 2003

Posts: 4441

Followers: 10

Kudos [?]: 82 [0], given: 0

[#permalink]

05 Jul 2004, 18:54

Is it:
1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9
_________________

Best Regards,

Paul

GMAT Club Legend

Joined: 15 Dec 2003

Posts: 4441

Followers: 10

Kudos [?]: 82 [0], given: 0

[#permalink]

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.
_________________

Best Regards,

Paul

[#permalink] 05 Jul 2004, 19:22

		Similar topics	Author	Replies	Last post
Similar Topics:		A and B alternately toss a coin. The first one to turn up a	araspai	3	09 Oct 2003, 21:43
		A and B alternately toss a coin. The first one to turn up a	mbassmbass04	6	11 Apr 2005, 05:32
		For one toss of a certain coin, the probability that the	zoom	4	10 Sep 2005, 13:54
		For one toss of a certain coin, the probability that the	john2005	7	22 May 2006, 15:13
	4	A man alternately tosses a coin and throws a die beginning	pradhan	9	25 Dec 2009, 23:34

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

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.

GMAT Courses

Admissions Consulting

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

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

Main navigation

GMAT Resources

Partners

MBA Resources