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M20-29

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M20-29  [#permalink]

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New post 16 Sep 2014, 01:09
1
7
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

72% (01:21) correct 28% (01:33) wrong based on 138 sessions

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Re M20-29  [#permalink]

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New post 16 Sep 2014, 01:09
Official Solution:

If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards randomly extracted from the bin will be greater than 7?

A. \(\frac{1}{2}\)
B. \(\frac{3}{4}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)

In only one outcome, when cards with 2 and 4 are selected, the sum will not be greater than 7. The probability of drawing 2 and 4 \(= \frac{1}{5}*\frac{1}{4} + \frac{1}{5}*\frac{1}{4} = \frac{1}{10}\) (2 and 4 can be drawn in a different order). The answer to the question is \(1 - \frac{1}{10} = \frac{9}{10}\).

Answer: D
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New post 06 Apr 2015, 09:27
Hi Bunuel,

I tried to solve the question with another approach but it seems to be wrong.

I tried to count the no of cases
The no of cases whose sum is greater than 7 when 2 is included : (2,6) , (2,8) , (2,10) = 3
The no of cases whose sum is greater than 7 when 4 is included : (4,6) , (4,8) , (4,10) = 3
The no of cases whose sum is greater than 7 when 6 is included : (6,4), (6,8) , (6,10) = 3
The no of cases whose sum is greater than 7 when 8 is included : (8,2), (8,4), (8,6), (8,10) = 4
The no of cases whose sum is greater than 7 when 10 is included :(10,2),(10,4),(10,6),(10,8)= 4

So total cases = 3+3+3+4+4 = 17.

And total number of outcomes can be calculated as 5C2.

Please help where I am going wrong ? Is 2,8 and 8,2 one case ?

Thanks
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New post 06 Apr 2015, 09:45
karant wrote:
Hi Bunuel,

I tried to solve the question with another approach but it seems to be wrong.

I tried to count the no of cases
The no of cases whose sum is greater than 7 when 2 is included : (2,6) , (2,8) , (2,10) = 3
The no of cases whose sum is greater than 7 when 4 is included : (4,6) , (4,8) , (4,10) = 3
The no of cases whose sum is greater than 7 when 6 is included : (6,4), (6,8) , (6,10) = 3
The no of cases whose sum is greater than 7 when 8 is included : (8,2), (8,4), (8,6), (8,10) = 4
The no of cases whose sum is greater than 7 when 10 is included :(10,2),(10,4),(10,6),(10,8)= 4

So total cases = 3+3+3+4+4 = 17.

And total number of outcomes can be calculated as 5C2.

Please help where I am going wrong ? Is 2,8 and 8,2 one case ?

Thanks


1. You are missing (6, 2) case there. This will make the number of favorable outcomes equal to 18.
2. The total number of outcomes is 5*4 = 20, not 10. 5 options for first number and 4 options for the second number.

Thus, P = 18/20 = 9/10.

Hope it's clear.
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Re: M20-29  [#permalink]

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New post 17 Dec 2016, 05:22
Bunuel wrote:
Official Solution:

If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards randomly extracted from the bin will be greater than 7?

A. \(\frac{1}{2}\)
B. \(\frac{3}{4}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)

In only one outcome, when cards with 2 and 4 are selected, the sum will not be greater than 7. The probability of drawing 2 and 4 \(= \frac{1}{5}*\frac{1}{4} + \frac{1}{5}*\frac{1}{4} = \frac{1}{10}\) (2 and 4 can be drawn in a different order). The answer to the question is \(1 - \frac{1}{10} = \frac{9}{10}\).

Answer: D




Hi,

What I tried was
there are 10 ways to choose 2 cards from 5. Ot these choices only 2 gives sum less than 7.
2 & 4 or 4&2

so probability of those choices is 2/10 = 1/5
hence probability of getting sum > 7 is 1-1/5 = 4/5

Can you please tell me, where I'm going wrong?
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Re: M20-29  [#permalink]

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New post 17 Dec 2016, 05:26
sidoknowia wrote:
Bunuel wrote:
Official Solution:

If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards randomly extracted from the bin will be greater than 7?

A. \(\frac{1}{2}\)
B. \(\frac{3}{4}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)

In only one outcome, when cards with 2 and 4 are selected, the sum will not be greater than 7. The probability of drawing 2 and 4 \(= \frac{1}{5}*\frac{1}{4} + \frac{1}{5}*\frac{1}{4} = \frac{1}{10}\) (2 and 4 can be drawn in a different order). The answer to the question is \(1 - \frac{1}{10} = \frac{9}{10}\).

Answer: D




Hi,

What I tried was
there are 10 ways to choose 2 cards from 5. Ot these choices only 2 gives sum less than 7.
2 & 4 or 4&2

so probability of those choices is 2/10 = 1/5
hence probability of getting sum > 7 is 1-1/5 = 4/5

Can you please tell me, where I'm going wrong?


In the first case you count groups of two so in the second case you also should count in groups of two. So, 2 and 4 is only one group. 1 - 1/10 = 9/10.
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Re: M20-29  [#permalink]

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New post 27 May 2017, 06:45
Bunuel wrote:
If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards randomly extracted from the bin will be greater than 7?

A. \(\frac{1}{2}\)
B. \(\frac{3}{4}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)


Probability = no of favourable cases / total number of cases.

Let's find the total number of cases = 5 cards and we need to pick any two - then 5C2 = 10.

Let's find the favourable cases that are greater than 7..

consider with starting with 2..
(2,6)(2,8)(2,10)
With 4
(4,6)(4,8)(4,10)
with 6
(6,8) (6,10)
with 8
(8,10).... Total 9 cases..

Probability = 9 / 10...

Hope this simple and clear...:)
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Re: M20-29  [#permalink]

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New post 01 Jul 2017, 02:11
I think this way is also possible.

Two number picked
5C2=10

Now, we know that the only combination out of 10 that does not exceed 7 is (4,2).

Hence the answer. 9/10
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New post 08 Aug 2017, 08:54
Hi,

Can we do:

Total number of ways= 5!/(3!x2!) = choose 2 among 5 of the number unordered = 10
Only 1 combination unordered not favorable (2;4)

Not favorable probability = 1 /10

Favorable case = 1 - 1/10 = 9/10

?
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New post 13 Sep 2017, 18:20
Tpral wrote:
Hi,

Can we do:

Total number of ways= 5!/(3!x2!) = choose 2 among 5 of the number unordered = 10
Only 1 combination unordered not favorable (2;4)

Not favorable probability = 1 /10

Favorable case = 1 - 1/10 = 9/10

?


This is the case of Permutation, not only Combination. You use combination formula here, but fortunately you reach the same answer.

As Bunuel explanation, [2,4] and [4,2] counted as two different cases.
Total possible outcomes : 5P2, which is 20.
Probability then become =\(1 - \frac{2}{20} = \frac{18}{20} = \frac{9}{10}\)

Bunuel please correct my approach ya.
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New post 23 Oct 2017, 02:44
+1 for D. Of the 5 numbers mentioned , only a combination of 2 & 4 will give a sum less than 7. The required probability is 1-(1/5c2). This comes out to 9/10. Hence option D.
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New post 08 Aug 2019, 18:56
Bunuel Is 4C2 +3 correct for favorable outcomes?
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New post 13 Nov 2019, 23:05
Is the word numbered a hint to use permutation?

Can any expert tell me when to decide whether order is imp or not?

Explaining a few cases would help me a lot.


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New post 14 Nov 2019, 06:25
axezcole wrote:
Is the word numbered a hint to use permutation?

Can any expert tell me when to decide whether order is imp or not?

Explaining a few cases would help me a lot.


Bunuel, chetan2u, GMATPrepNow, MartyTargetTestPrep, ScottTargetTestPrep, VeritasKarishma


Here is a post on our blog explaining when order matters:
https://www.veritasprep.com/blog/2013/0 ... er-matter/
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New post 15 Nov 2019, 13:20
axezcole wrote:
Is the word numbered a hint to use permutation?

Can any expert tell me when to decide whether order is imp or not?

Explaining a few cases would help me a lot.


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Since there's no replacement in this problem, it does not matter which method you use for this question; it can be considered as either a combination problem or a permutation problem (hence the same answer for both methods).

If you interpret the experiment as drawing the first card from a total of five and then drawing another card from the remaining four, the problem becomes a permuation problem. In this interpretation, drawing a 2 first and an 8 next is different from the outcome of drawing an 8 first and a 2 next.

If, on the other hand, you interpret the experiment as drawing two cards at the same time from the total of five, then the question becomes a combination problem. When the two cards you draw are a 2 and an 8, that's a single outcome as you draw the two cards at the same time.

In the first interpretation, the total number of outcomes and the number of favorable outcomes are multiplied by two; since each outcome can happen in exactly two ways (such as (2, 8) or (8, 2)). That's why the probability we need to calculate will be the same regardless of whether we use the first interpretation or the second interpretation.

Had there been replacement, it would have not worked out like above. With replacement, you also need to take into account outcomes of (2, 2), (4, 4) etc.; thus you could no longer use the combination method.

It is also interesting to think about how would each method work if you picked not two but three cards.
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New post 25 Nov 2019, 05:51
msk0657 wrote:
Bunuel wrote:
If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards randomly extracted from the bin will be greater than 7?

A. \(\frac{1}{2}\)
B. \(\frac{3}{4}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)


Probability = no of favourable cases / total number of cases.

Let's find the total number of cases = 5 cards and we need to pick any two - then 5C2 = 10.

Let's find the favourable cases that are greater than 7..

consider with starting with 2..
(2,6)(2,8)(2,10)
With 4
(4,6)(4,8)(4,10)
with 6
(6,8) (6,10)
with 8
(8,10).... Total 9 cases..

Probability = 9 / 10...

Hope this simple and clear...:)



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New post 25 Nov 2019, 05:56
axezcole wrote:
msk0657 wrote:
Bunuel wrote:
If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards randomly extracted from the bin will be greater than 7?

A. \(\frac{1}{2}\)
B. \(\frac{3}{4}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)


Probability = no of favourable cases / total number of cases.

Let's find the total number of cases = 5 cards and we need to pick any two - then 5C2 = 10.

Let's find the favourable cases that are greater than 7..

consider with starting with 2..
(2,6)(2,8)(2,10)
With 4
(4,6)(4,8)(4,10)
with 6
(6,8) (6,10)
with 8
(8,10).... Total 9 cases..

Probability = 9 / 10...

Hope this simple and clear...:)



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Re: M20-29  [#permalink]

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New post 26 Nov 2019, 00:57
total possible card draws = 5*4 ; 20
and pairs sum>7 ( 2,6) ( 2,8) ( 2,10) ( 4,6) ( 4,8) ( 4,10) ( 6,2) ( 6,4) ( 6,8) ( 6,10) ( 8,2) ( 8,4) ( 8,6) ( 8,10) ( 10,2) (10,4)(10,6) (10,8)
total = 18
P = 18/20 ; 9/10
IMO D


Bunuel wrote:
If a bin contains five cards numbered 2, 4, 6, 8, and 10, what is the probability that the sum of the numbers on two cards randomly extracted from the bin will be greater than 7?

A. \(\frac{1}{2}\)
B. \(\frac{3}{4}\)
C. \(\frac{4}{5}\)
D. \(\frac{9}{10}\)
E. \(\frac{19}{20}\)
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Re: M20-29   [#permalink] 26 Nov 2019, 00:57
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