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Odissia Just gettin' started!
Joined: 01 Oct 2007 Posts: 2
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Location: Chicago
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Posted: Thu Oct 04, 2007 6:43 pm Post subject: probability |
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Hi everybody,
here is one question:
Three cards are chosen at rendom from a desk without replacement. what is the probability of getting a jack, a tan and a nine in order?
1. 8\16575
2. 1\2197
3. 6\35152
4. none of the above
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Posted: Fri Oct 05, 2007 6:13 am Post subject: |
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I haven't come to the right answer yet, but here's my approach:
The answer will be a sum of the probabilities of 4 different scenarios. P(J and T and 9 are all independent) + P(J and T are dependent but not 9) + P(T and 9 are dependent but not J) + P(J and T and 9 are all dependent)
The jack you pick out is not a Tan and the Tan you pick out is not a 9 (4/56)x(28/55)x(4/54)
The Jack you pick out is a Tan but the Tan you pick out is not a 9 (4/56)x(27/55)x(4/54)
The Jack you pick out is not a Tan but the Tan you pick out is a 9 (4/56)x(28/55)x(3/54)
The Jack you pick out is a Tan and the Tan you pick out is a 9 (4/56)x(27/55)x(3/54)
Then you sum all of that up. [4/(56x55x54)] x [(28x4) + (27x4) + (28x3) + (27x3)]
(1/41580) x (112+108+84+81) = 385/41580 |
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Leonard C Just gettin' started!
Joined: 16 Jul 2007 Posts: 29
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Posted: Sun Oct 07, 2007 3:32 am Post subject: |
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You can do this simply by:
(4/52) x (4/51) x (4/50) = 8/16,575.
You start with 52 cards, but since there is no replacement, your sample population decreases by one. |
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sonibubu Just gettin' started!
Joined: 21 Sep 2007 Posts: 22
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Test Date: March 7, 2008 Target GMAT Score: 750
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Posted: Thu Oct 11, 2007 9:13 am Post subject: |
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I agree with Leonard. The whole "tan" thing threw me off, I was wondering if there are tan colored cards  |
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joshi.komal Rising GMAT Star
Joined: 25 Sep 2007 Posts: 76
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Posted: Sun Oct 14, 2007 8:59 am Post subject: |
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Well..even I am scared of probability
But Now I have started getting the feel about it.
The best way when you get such problems where you are taking out something and not putting back then the number at the bottom will decrease by one which is your sample space. In this case it is the total no of cards or you may even have balls in a bag.
Next is how many jacks out of the total are there in the deck it is 4 so your probability for the first time will be: you will get any 4 out of total 52.
So probability of getting jack is 4/52.
now you have to ppick ten but the number of cards have reduced by 1.
So probability will 4 out 51 i.e 4/51.
Similarly the next will Nine 4/50.
so the total probability will be a product of these.
4/52*4/51*4/50
Thanks
Komal |
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Odissia Just gettin' started!
Joined: 01 Oct 2007 Posts: 2
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Location: Chicago
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Posted: Sun Oct 14, 2007 12:45 pm Post subject: good job! |
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they are really good explanations! Thanks for all of you.
I downlowded some materials from this web site about probability which are very helpful.
Odissia. |
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