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The Full House Casino is running a new promotion. Each

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Intern
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The Full House Casino is running a new promotion. Each [#permalink]

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27 Sep 2012, 11:41
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The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

A. 1/336
B. 1/120
C. 1/56
D. 1/720
E. 1/1440
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Sep 2012, 11:46, edited 1 time in total.
Renamed the topic and moved to PS forum.

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Re: The Full House Casino is running a new promotion. Each [#permalink]

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27 Sep 2012, 11:48
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joesamson wrote:
The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

A. 1/336
B. 1/120
C. 1/56
D. 1/720
E. 1/1440

P=P(ace in the first draw)*P(ace in the second draw)*P(ace in the third draw)=3/8*2/7*1/6=1/56.

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Re: The Full House Casino is running a new promotion. Each [#permalink]

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28 Sep 2012, 12:47
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The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

There is only one result that results in a win: receiving three aces.

Since the order of arrangement does not matter, the number of possible ways to receive 3 cards is a combination problem.

The number of combinations of n objects taken r at a time is

C(n,r) = n!/(r!(n-r)!)

C(8,3) = 8!/(3!(8-3)!)
C(8,3) = 8!/(3!(5!))
C(8,3) = 40320/(6(120))
C(8,3) = 40320/720
C(8,3) = 56

The number of possible outcomes is 56.

Thus, the probability of being dealt 3 aces is 1/56.

HENCE C

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Re: The Full House Casino is running a new promotion. Each [#permalink]

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29 Sep 2012, 23:42
Bunuel wrote:
joesamson wrote:
The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

A. 1/336
B. 1/120
C. 1/56
D. 1/720
E. 1/1440

P=P(ace in the first draw)*P(ace in the second draw)*P(ace in the third draw)=3/8*2/7*1/6=1/56.

Bunuel.. If question wud ask ...with replacement then??

then solution wud b like this..

3/8*3/8*3/8= 27/512..rite??
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Re: The Full House Casino is running a new promotion. Each [#permalink]

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01 Oct 2012, 05:49
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Expert's post
sanjoo wrote:
Bunuel wrote:
joesamson wrote:
The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

A. 1/336
B. 1/120
C. 1/56
D. 1/720
E. 1/1440

P=P(ace in the first draw)*P(ace in the second draw)*P(ace in the third draw)=3/8*2/7*1/6=1/56.

Bunuel.. If question wud ask ...with replacement then??

then solution wud b like this..

3/8*3/8*3/8= 27/512..rite??

Yes, since in this case, each time we would be picking from 8 cards with 3 aces.
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Re: The Full House Casino is running a new promotion. Each [#permalink]

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01 Oct 2012, 08:21
ok got that ..Thanks bunuel:)
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Re: The Full House Casino is running a new promotion. Each [#permalink]

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11 Oct 2012, 03:15
Don't we have to consider the order of the Aces?

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Re: The Full House Casino is running a new promotion. Each [#permalink]

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11 Oct 2012, 03:27
tanujab wrote:
Don't we have to consider the order of the Aces?

All Aces are Aces, order doesn't matter.
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Re: The Full House Casino is running a new promotion. Each [#permalink]

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11 Oct 2012, 03:35
Thanks Eva for the clarification

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Re: The Full House Casino is running a new promotion. Each [#permalink]

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02 Nov 2014, 22:17
Hello from the GMAT Club BumpBot!

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Re: The Full House Casino is running a new promotion. Each [#permalink]

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14 Jan 2015, 09:42
Or we can do:

8!/3!*5!

In terms of calculation we can skip a fair bit like this:
1*2*3*4*5*6*7*8/ (1*2*3)(1*2*3*4*5). So, 5! is included in 8! and is simplified, like this:

6*7*8/1*2*3 --> 6=2*3, so this also gets simplified.

We end up with 7*8= 56 and 1. So, ANS C

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Re: The Full House Casino is running a new promotion. Each [#permalink]

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13 Jun 2016, 06:34
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The Full House Casino is running a new promotion. Each   [#permalink] 13 Jun 2016, 06:34
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