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If the probability of pulling two cards from a stack of uniquely numbe

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If the probability of pulling two cards from a stack of uniquely numbe [#permalink]

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New post 27 Mar 2017, 04:47
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Question Stats:

82% (02:25) correct 18% (03:44) wrong based on 41 sessions

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If the probability 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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Re: If the probability of pulling two cards from a stack of uniquely numbe [#permalink]

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New post 27 Mar 2017, 06:50
Bunuel wrote:
If the probability 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


Two possibilities: Out of total(n) cards,
1. 1st is a 5 and 2nd is a 6
2. 1st is a 6 and 2nd is a 5.

Total probability - > (1/n) * (1/n-1) * 2 = 0.1. Solving the equation we get n=5.

Hence C.

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Re: If the probability of pulling two cards from a stack of uniquely numbe [#permalink]

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New post 27 Mar 2017, 13:54
You could try process elimination here as well.

getting the five and six is 0.10

Let's try answer C 5 = (FS)+ (SF) = (1/5 x 1/4) + (1/5 x 1/4) = 1/10

Therefore the answer is C
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Re: If the probability of pulling two cards from a stack of uniquely numbe [#permalink]

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New post 28 Mar 2017, 21:01
1
Hi All,

This question can be solved by TESTing THE ANSWERS.

We're told that the probability of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the '5' and '6' is 0.10... We're asked for the total number of cards in the stack.

To start, it's worth noting that .1 is a relatively 'nice' decimal. Probability questions almost always involve multiplying fractions - and the results are often NOT 'nice' decimals. This means that we're likely dealing with a total that can be converted into 'nice' fractions (for example 1/4 is 'nice', while 1/11 is NOT).

Let's TEST Answer B: 4 cards

Pulling the '5' or the '6' on the first try = 2/4
Pulling the other card on the second try = 1/3
Probability of pulling the two cards we want = (2/4)(1/3) = 2/12 = 1/6. This is NOT a match for what we were told (it's supposed to be .1 = 1/10).

Let's TEST Answer C: 5 cards

Pulling the '5' or the '6' on the first try = 2/5
Pulling the other card on the second try = 1/4
Probability of pulling the two cards we want = (2/5)(1/4) = 2/20 = 1/10. This IS an exact match for what we were told, so this MUST be the answer.

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Re: If the probability of pulling two cards from a stack of uniquely numbe [#permalink]

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New post 29 Mar 2017, 10:03
1
Bunuel wrote:
If the probability 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 getting the 5 and the 6 is the probability of getting the 5 followed by the 6 AND the probability of getting the 6 followed by the 5.

If we let n = the number of cards in the deck, the probability of getting the 5 followed by the 6 is 1/n x 1/(n - 1), and this also will be the probability of the 6 followed by the 5. Since we are given that the probability of selecting the 5 and the 6 is 0.1 = 1/10, we can create the following equation:

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

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

n^2 - n = 20

n^2 - n - 20 = 0

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

n = 5 or n = -4

Since n can’t be negative, n = 5.

Answer: C
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Re: If the probability of pulling two cards from a stack of uniquely numbe   [#permalink] 29 Mar 2017, 10:03
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