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The Full House Casino is running a new promotion. Each [#permalink]
 27 Sep 2012, 12:41
Question Stats:
78% (01:44) correct
21% (03:26) wrong based on 4 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? A. 1/336 B. 1/120 C. 1/56 D. 1/720 E. 1/1440
Last edited by Bunuel on 27 Sep 2012, 12:46, 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]
27 Sep 2012, 12:48
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.
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Re: The Full House Casino is running a new promotion. Each [#permalink]
28 Sep 2012, 13:47
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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]
30 Sep 2012, 00:42
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??
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Re: The Full House Casino is running a new promotion. Each [#permalink]
01 Oct 2012, 06:49
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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.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
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Re: The Full House Casino is running a new promotion. Each [#permalink]
01 Oct 2012, 09:21
ok got that ..Thanks bunuel:)
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Re: The Full House Casino is running a new promotion. Each [#permalink]
11 Oct 2012, 04:15
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]
11 Oct 2012, 04:27
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]
11 Oct 2012, 04:35
Thanks Eva for the clarification 
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Re: The Full House Casino is running a new promotion. Each
[#permalink]
11 Oct 2012, 04:35
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