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
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Difficult Math Problem #75 - Probability
This topic has expert replies
-
mukul
- Junior | Next Rank: 30 Posts
- Posts: 26
- Joined: Wed Oct 11, 2006 11:02 pm
- Location: Tuck School of Business, U of Dartmouth
- Thanked: 4 times
- Followed by:8 members
- GMAT Score:770
I think its 1/56
ways of choosing three aces from the deck is 1....there are only three aces...so theres only one way you can have all three.
Now ways of choosing 3 cards from a deck of 8 cards is 8C3 = 56
so answer is 1/56
hope that helps..
ways of choosing three aces from the deck is 1....there are only three aces...so theres only one way you can have all three.
Now ways of choosing 3 cards from a deck of 8 cards is 8C3 = 56
so answer is 1/56
hope that helps..
GMAT/MBA Expert
-
beatthegmat
- Site Admin
- Posts: 6773
- Joined: Mon Feb 13, 2006 8:30 am
- Location: Los Angeles, CA
- Thanked: 1249 times
- Followed by:994 members
Nice to see you still posting, Mukul! You've got a lot of knowledge to share with this community, along with some unquestionable street credibility with your GMAT score!
Beat The GMAT | The MBA Social Network
Community Management Team
Research Top GMAT Prep Courses:
https://www.beatthegmat.com/gmat-prep-courses
Research The World's Top MBA Programs:
https://www.beatthegmat.com/mba/school
Community Management Team
Research Top GMAT Prep Courses:
https://www.beatthegmat.com/gmat-prep-courses
Research The World's Top MBA Programs:
https://www.beatthegmat.com/mba/school
-
800guy
- Master | Next Rank: 500 Posts
- Posts: 354
- Joined: Tue Jun 27, 2006 9:20 pm
- Thanked: 11 times
- Followed by:5 members
OA:
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.
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.
-
jasminekaur280
- Junior | Next Rank: 30 Posts
- Posts: 11
- Joined: Sun May 31, 2020 9:47 pm
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.
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.
-
quichopipe
- Junior | Next Rank: 30 Posts
- Posts: 12
- Joined: Wed Apr 08, 2020 12:42 pm
Let's remember that probability simply says (number of favorable outcomes)/(total possible outcomes).
In this case, to win the trip, a player must get all three aces. Since the player receives 3 cards and there are only 3 aces, there is only ONE way that the player gets all the aces...ace on the first deal, ace on the second deal, and ace on the third deal. So our numerator is simply 1.
Lets' calculate our denominator. We have 8 cards and we want to get 3 of them. Therefore, our denominator is 8C3 = 56
1/56
Answer choice C
In this case, to win the trip, a player must get all three aces. Since the player receives 3 cards and there are only 3 aces, there is only ONE way that the player gets all the aces...ace on the first deal, ace on the second deal, and ace on the third deal. So our numerator is simply 1.
Lets' calculate our denominator. We have 8 cards and we want to get 3 of them. Therefore, our denominator is 8C3 = 56
1/56
Answer choice C
