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If the chance of pulling two cards from a stack of uniquely numbered c

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duahsolo
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If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   10 Feb 2017, 03:07
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If the chance of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the five and six is 0.10, then how many cards are in the stack?

A) 2
B) 4
C) 5
D) 11
E) 12
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C
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chetan2u
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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   10 Feb 2017, 04:51
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duahsolo wrote:
If the chance of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the five and six is 0.10, then how many cards are in the stack?

A) 2
B) 4
C) 5
D) 11
E) 12



Hi,

Let the number of cards be n....
So getting 5 or 6 will be getting any TWO of n =\(\frac{2}{n}\)...
The remaining numbered card will now be choosen from n-1 card, so prob=\(\frac{1}{n-1}\)..

So the overall prob =\(\frac{2}{n}*\frac{1}{n-1}=0.10.............\frac{2}{n(n-1)}=0.10.........20=n(n-1)=5*4\)...
So n =5..
C
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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   10 Feb 2017, 07:18
chetan2u wrote:
duahsolo wrote:
If the chance of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the five and six is 0.10, then how many cards are in the stack?

A) 2
B) 4
C) 5
D) 11
E) 12



Hi,

Let the number of cards be n....
So getting 5 or 6 will be getting any TWO of n =\(\frac{2}{n}\)...
The remaining numbered card will now be choosen from n-1 card, so prob=\(\frac{1}{n-1}\)..

So the overall prob =\(\frac{2}{n}*\frac{1}{n-1}=0.10.............\frac{2}{n(n-1)}=0.10.........20=n(n-1)=5*4\)...
So n =5..
C


Ok try plugging it back in though... if there are 5 cards in the deck the chance of getting a 5 are 1/5 and then the chance of getting 6 out of the 4 cards remaining are 1/4 SO 1/5 * 1/4 does not equal to 1/10 it actually equals 1/125


HELP
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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   15 Feb 2017, 10:47
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duahsolo wrote:
If the chance of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the five and six is 0.10, then how many cards are in the stack?

A) 2
B) 4
C) 5
D) 11
E) 12


We are given that the probability of drawing a 5 and 6 from a deck of uniquely numbered cards, without replacement, is 0.10. If we let n = the total number of cards, we see that the probability of drawing either the 5 or the 6 on the first draw is 2/n. Since one of the desired cards was drawn on the first draw, we see that the probability of getting the other desired card on the second draw will be 1/n-1. Thus, we can create the following equation to determine n:

2/n x 1/(n-1) = 1/10

2/(n^2 - n) = 1/10

20 = n^2 - n

n^2 - n - 20 = 0

(n - 5)(n + 4) = 0

n = 5 or n = -4

Since n must be positive, n = 5.

Answer: C
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onamarif
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If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   19 Feb 2017, 00:19
Let the number of cards be x

\(2/x\) * \(1/(x-1)\)= \(1/10\) => \(2/x(x-1)\) = \(1/10\)
\(x^2 -x =20 => x^2 -x -20 =0\)
This gives x=5 and x=-4.
Since number of cards cannot be a negative quantity, 5 is the number of cards
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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   18 Mar 2017, 01:32
rishit1080 wrote:
chetan2u wrote:
duahsolo wrote:
If the chance of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the five and six is 0.10, then how many cards are in the stack?

A) 2
B) 4
C) 5
D) 11
E) 12



Hi,

Let the number of cards be n....
So getting 5 or 6 will be getting any TWO of n =\(\frac{2}{n}\)...
The remaining numbered card will now be choosen from n-1 card, so prob=\(\frac{1}{n-1}\)..


So the overall prob =\(\frac{2}{n}*\frac{1}{n-1}=0.10.............\frac{2}{n(n-1)}=0.10.........20=n(n-1)=5*4\)...
So n =5..
C


Ok try plugging it back in though... if there are 5 cards in the deck the chance of getting a 5 are 1/5 and then the chance of getting 6 out of the 4 cards remaining are 1/4 SO 1/5 * 1/4 does not equal to 1/10 it actually equals 1/20


HELP


i have followed the same method but i think the answer that you have mentioned should be 1/20 not 1/125
but can we do it again i mean as we got 5 on the first chance and 6 on the 2nd chance so same if we got 6 on the first and 5 on the 2nd then the total prob. will be 1/20+1/20=1/10 that is answer , i may be wrong still waiting for the correction . :roll:
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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   18 Mar 2017, 23:34
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We can solve this by trying out the answer choices.
Starting with "c"
Probability of 1st draw is 2/5
Probability of 2nd draw =1/4 (since one of the numbers has already been drawn out in the first attempt)
(2/5)*(1/4)=1/10

Thus correct option is "C"
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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   20 Mar 2017, 15:28
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Hi rishit1080,

In this question, the order of the two cards does NOT matter, so there are two ways to get the end result that we're after...

1) Get the 5 first and the 6 second.
2) Get the 6 first and the 5 second.

Thus, when we pull the first card, there are TWO options that fit what we're looking for (the 5 OR the 6) - and on the second card, there is ONE option that fits (whatever card we didn't pull first). With five total cards, the math would be....

(2/5)(1/4) = 2/20 = 1/10

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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink] If the chance of pulling two cards from a stack of uniquely numbered c   30 Apr 2024, 03:10
rishit1080 wrote:
chetan2u wrote:
duahsolo wrote:
If the chance of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the five and six is 0.10, then how many cards are in the stack?

A) 2
B) 4
C) 5
D) 11
E) 12

Hi,

Let the number of cards be n....
So getting 5 or 6 will be getting any TWO of n =\(\frac{2}{n}\)...
The remaining numbered card will now be choosen from n-1 card, so prob=\(\frac{1}{n-1}\)..

So the overall prob =\(\frac{2}{n}*\frac{1}{n-1}=0.10.............\frac{2}{n(n-1)}=0.10.........20=n(n-1)=5*4\)...
So n =5..
C

Ok try plugging it back in though... if there are 5 cards in the deck the chance of getting a 5 are 1/5 and then the chance of getting 6 out of the 4 cards remaining are 1/4 SO 1/5 * 1/4 does not equal to 1/10 it actually equals 1/125


HELP

­Hi Rishit1080,
1/4*1/5=1/20. You should also consider the ordering here ie (5,6) and (6,5) are two different cases hence 2/20=1/10. Hope this helps!­
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Re: If the chance of pulling two cards from a stack of uniquely numbered c [#permalink]
30 Apr 2024, 03:10
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