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Plug in the values: \(P = C^3_2*(\frac{1}{4})^2*(1-\frac{1}{4})^{(3-2)}\) \(C^3_2*(\frac{1}{4})^2*(\frac{3}{4})^{1}\) \(3*\frac{1}{16}*\frac{3}{4}=\frac{9}{64}\) _________________
Re: Easy Combination, but need help! [#permalink]
26 Apr 2011, 13:20
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?
Re: Easy Combination, but need help! [#permalink]
26 Apr 2011, 13:31
johnnybear wrote:
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?
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 _________________
Re: Easy Combination, but need help! [#permalink]
26 Apr 2011, 17:07
Expert's post
johnnybear wrote:
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?
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
Plug in the values: \(P = C^3_2*(\frac{1}{4})^2*(1-\frac{1}{4})^{(3-2)}\) \(C^3_2*(\frac{1}{4})^2*(\frac{3}{4})^{1}\) \(3*\frac{1}{16}*\frac{3}{4}=\frac{9}{64}\)
Is someone supposed to memorize that for the test? Are there any guides to learning probability as a process? _________________
"What we obtain too cheap, we esteem too lightly." -Thomas Paine
Re: Easy Combination, but need help! [#permalink]
27 Apr 2011, 00:31
havok wrote:
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.
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. _________________
Re: Easy Combination, but need help! [#permalink]
01 May 2011, 02:46
fluke & VeritasPrepKarishma wrote:
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
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 _________________
WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.
Re: Easy Combination, but need help! [#permalink]
01 May 2011, 17:01
Expert's post
Warlock007 wrote:
fluke & VeritasPrepKarishma wrote:
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
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? _________________
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