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The Full House Casino is running a new promotion. Each [#permalink]
27 Sep 2012, 11:41

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

77% (01:52) correct
23% (01:51) wrong based on 121 sessions

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?

Re: The Full House Casino is running a new promotion. Each [#permalink]
27 Sep 2012, 11:48

Expert's post

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.

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

Answer: C.

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

then solution wud b like this..

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

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Re: The Full House Casino is running a new promotion. Each [#permalink]
01 Oct 2012, 05:49

1

This post received KUDOS

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.

Answer: C.

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. _________________

Re: The Full House Casino is running a new promotion. Each [#permalink]
02 Nov 2014, 22:17

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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Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...