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Difficult Math Problem #75 - Probability

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Difficult Math Problem #75 - Probability

by 800guy » 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
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answer if 1/56

by mukul » 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

so answer is 1/56

hope that helps..
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by beatthegmat » 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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OA

by 800guy » 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.
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by thankont » 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).
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Re: Difficult Math Problem #75 - Probability

by jasminekaur280 » Wed Jun 03, 2020 12:27 am
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.
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Re: Difficult Math Problem #75 - Probability

by quichopipe » Wed Jun 03, 2020 1:01 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
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