Easy Combination, but need help!

Question

Hey guys, new to the forum here, I hope to ask a lot of questions and also answer many questions that may fall into my strengths.

I have a supposedly easy combination question:

3 cards are drawn at random from a deck of 52 cards WITH replacement. How many ways can I get exactly 2 hearts?

fluke · Answer

Welcome to the forum. Hope you learn and share a lot!!! All the best.

Probability of an event occurring k times in n-time sequence could be expressed as:

P = C^n_k*P(Favorable)^k*(1-P(Favorable))^{(n-k)}

n=3 
 k=2 
 P(Heart)=\frac{13}{52}=\frac{1}{4} 
 P(Non-Heart)=1-P(Heart)=1-\frac{1}{4}=\frac{3}{4}

Plug in the values:
 P = C^3_2*(\frac{1}{4})^2*(1-\frac{1}{4})^{(3-2)} 
 C^3_2*(\frac{1}{4})^2*(\frac{3}{4})^{1} 
 3*\frac{1}{16}*\frac{3}{4}=\frac{9}{64}

johnnybear · Answer

Thanks for the welcome and for the reply.

The probability part was helpful, but I am still unable to figure out how many combinations there are exactly. Could you maybe help me with that part as well?

fluke · Answer

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

KarishmaB · Answer

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}

havok · Answer

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

fluke · Answer

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.

Warlock007 · Answer

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

KarishmaB · Answer

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?

Easy Combination, but need help!

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