Probability!!!!!!!!!!!!!

[stalwar85]

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

8/33

62/165

17/33

103/165

25/33

Probability isn't my strong point, and I am having issues with this problem for sure. I just need some help on how to lay out this problem or any probability problems.
Thanks a lot!

[tim]

i'll be honest with you, i just read MGMAT's explanation for this one and it's beautiful. We're happy to help you with this one, but you're going to have to provide some more information about what you don't understand that is not covered by the official explanation..

[stalwar85]

I just dont understand probability as a whole and getting this question is just very difficult to solve. I guess if I could just get a step by step layout. Like I have read over the explanation over and over and I mind just gets more and more confused instead of being cleared up.

[tim]

again, you're going to have to be more specific. i suspect any of our instructors who answered this question would say almost exactly what the official solution says, which wouldn't be of much help to you. :) where are you stuck in the solution? that might help..

[memav7]

In this question it is easier to find out the chances of getting 4 card without any couple of them with the same value. This can be deducted from 1 to get the probability of at least one couple.

Card 1: can be any number at any suit. So we can chose 12 out of 12.
Card 2: can be any card from the remaining 11 besides the card, which we already hold its number. So it is 10 out of 11.
Card 3: can be any card from the remaining 10 besides the 2 cards, which we already hold the numbers. So it is 8 out of 10.
Card 4: the same idea, only now we have to deduct 3 invalid cards. So it is 6 out of 9.

Probability of no couples = 1*(10/11)*(8/10)*(6/9) = 16/33.
Therefore the probability of at least one couple = 1 - 16/33 = 17/33

Option C.

Hope its correct.

Thanks

[RonPurewal]

In this question it is easier to find out the chances of getting 4 card without any couple of them with the same value. This can be deducted from 1 to get the probability of at least one couple.

Card 1: can be any number at any suit. So we can chose 12 out of 12.
Card 2: can be any card from the remaining 11 besides the card, which we already hold its number. So it is 10 out of 11.
Card 3: can be any card from the remaining 10 besides the 2 cards, which we already hold the numbers. So it is 8 out of 10.
Card 4: the same idea, only now we have to deduct 3 invalid cards. So it is 6 out of 9.

Probability of no couples = 1*(10/11)*(8/10)*(6/9) = 16/33.
Therefore the probability of at least one couple = 1 - 16/33 = 17/33

Option C.

Hope its correct.

Thanks

yes. nicely done.