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## Difficult Math Problem #75 - Probability

This topic has 1 expert reply and 3 member replies
800guy Master | Next Rank: 500 Posts
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#### Difficult Math Problem #75 - Probability

Wed Dec 13, 2006 4:25 pm
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

thankont Senior | Next Rank: 100 Posts
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Fri Dec 15, 2006 12:23 pm
One other approach is this one ( 3 /8 ).( 2/7 ).( 1/6 ) = 1/56 (since the question is without replacent).

800guy Master | Next Rank: 500 Posts
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Fri Dec 15, 2006 9:30 am
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.

mukul Junior | Next Rank: 30 Posts
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Thu Dec 14, 2006 4:27 am
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

hope that helps..

### GMAT/MBA Expert

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Thu Dec 14, 2006 5:16 pm
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!

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