Tough probability question

brijesh wrote:

gmatrant wrote:The question says "he turns over 4 cards" - which means picking 4 cards out of 12. Why would it be a permutation rather than just combination.

Quite don't agree. Please explain.

Hi!

Jim is right here, the cards (6 each) are from different suit- u can consider the other features, say color of the two suit may be diffrent (though not mentioned but otherwise the word '2 suit' does`t make sense).

even in your calculation u have used the value of 12P4 (12*11*10*9)- for calculating the total no of ways.

hope this make sense !

I posted this to gmatclub, so I might as well paste an edited version here:

Whether you pick the four cards simultaneously, or pick them one at a time (without replacement) doesn't actually matter if you are finding a probability; the two situations are mathematically identical. You can see this intuitively by thinking of taking hold of four cards in the deck first. If you take them all out at the same time, or if there is a nanosecond between your removing each, why would the probability that you get a pair be affected? It won't be, of course.

You can see that either perspective will give you the same answer, though it's easier to illustrate with a simpler example. Say you have 3 red marbles and 4 blue marbles in a bag, and you pick two (either simultaneously, or one at a time without replacement - it's the same thing), and you want to find the probability of picking two red marbles. If we look at the problem as though we are picking marbles one at a time, we have 3*2 ways of picking two reds, and 7*6 ways of picking two marbles, so the probability would be 3*2/7*6 = 1/7. If we look at the problem as though we're picking two marbles simultaneously, we have 3C2 ways of picking two red marbles and 7C2 ways of picking two marbles, so the probability would be 3C2/7C2 = 1/7. So when you stick your two hands in the bag and grab two marbles, it doesn't matter if you lift your two hands out at the same time, or take them out one at a time; the probability you pick two reds is the same.

Note though that you need to be consistent in the calculation - if you assume order matters when you calculate the numerator, you must also assume order matters when you calculate the denominator. But in any *probability* problem involving selecting cards (without replacement), or selecting marbles (without replacement), or flipping coins, etc, there is no 'correct' answer to the question 'does order matter?' You can pretend it does matter, or pretend it doesn't, and either way you'll get the right answer as long as you calculate the numerator and denominator in the same way. In counting problems, on the other hand, of course it's crucial to determine whether order matters.