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

Question

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

chetan2u · Answer

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

rishit1080 · Answer

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

JeffTargetTestPrep · Answer

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

onamarif · Answer

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

nks2611 · Answer

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:

scorpion740 · Answer

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"

EMPOWERgmatRichC · Answer

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

GMAT assassins aren't born, they're made,
Rich

SirSanguine · Answer

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!

luisdicampo · Answer

Deconstructing the Question

We draw 2 cards without replacement from a stack of n uniquely numbered cards.

We want the probability of getting exactly the cards 5 and 6.

Total ways to choose 2 cards:

\binom{n}{2} = \frac{n(n-1)}{2}

Only 1 favorable pair: {5,6}.

So the probability is:

\frac{1}{\binom{n}{2}} = 0.10

Step-by-step

\frac{1}{\frac{n(n-1)}{2}} = 0.10

\frac{2}{n(n-1)} = 0.10

Multiply both sides by  n(n-1) :

2 = 0.10 \cdot n(n-1)

Divide by 0.10:

20 = n(n-1)

n^2 - n - 20 = 0

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

Reject negative solution.

Answer: C

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

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

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