Easy Combination, but need help!

Ah!! I see!!

\(C^3_2*(13)^2*39\)

13- Number of ways you can pick hearts
39- Number of ways you can pick others
\(C^3_2\)- Number of ways to select the position of the 2 hearts out of 3

Or think of it this way:
2 of the 3 cards should be hearts. This can happen in 3 ways (i.e. 3C2): HHO, HOH, OHH
fluke has already explained that the probability of picking a Heart is 1/4 and of some other card is 3/4

Getting HHO: \((\frac{1}{4})(\frac{1}{4})(\frac{3}{4})\)
Getting HOH: \((\frac{1}{4})(\frac{3}{4})(\frac{1}{4})\)
Getting OHH: \((\frac{3}{4})(\frac{1}{4})(\frac{1}{4})\)

Total probability of getting exactly 2 hearts = \((\frac{3}{64})+(\frac{3}{64})+(\frac{3}{64})= \frac{9}{64}\)

Is someone supposed to memorize that for the test? Are there any guides to learning probability as a process?

It's good to be wary of the formula. This formula in particular is not as subtle as it seems. Practicing is the only way out for Probability & Combinatorics.

math-probability-87244.html
math-combinatorics-87345.html

Practice:
combined-probability-question-87294.html
combinations-permutations-and-probability-references-56486.html
hardest-area-questions-probability-and-combinations-101361.html

Good Books:
Veritas Prep: Combinatorics & Probability
MGMAT: Word Translations Guide

P.S.: Don't waste too much time on this. GMAT will ask only limited question from this type (maybe around 2). Better delve into it if you are through with other relatively easier topics with greater weight, such as inequality, ratio, percentage, sets, statistics.

Dear Fluke & karishma

do we really need to multiply \(C^3_2\) to \((13)^2*39\)
as in selection order dosent matter at all
all we need to ensure is that there should be two hearts and one any other cards
thats it

according to my knowledge answer should be \((13)^2*C^3_1^9\)
please clarify

Say you pick out a card from the deck and it is the 3 of Hearts. You return the card back. Then you pick out another one and it is the 4 of Hearts. You return it again. Then you pick out another card and that is the 5 of Spades.
Now imagine another situation. You pick out the first card and it is the 5 of Spades. You return it to the deck. You pick out another card and that is the 3 of Hearts. You put it back. Then you pick out another card and that is the 4 of Hearts.
Tell me, are the two cases the same? Isn't there a sequence involved? If you want to find the number of ways in which you can get the 3 of Hearts, the 4 of Hearts and the 5 of Spades, wouldn't you add these two cases?