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Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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7 00:00

Difficulty:   25% (medium)

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

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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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 $$1 - \frac{1}{10} = \frac{9}{10}$$.

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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.

Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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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.

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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Manager  B
Joined: 18 Jun 2016
Posts: 86
Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 700 Q49 V36 ### Show Tags

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 $$1 - \frac{1}{10} = \frac{9}{10}$$.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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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 $$1 - \frac{1}{10} = \frac{9}{10}$$.

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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Retired Moderator G
Joined: 26 Nov 2012
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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... Intern  B
Joined: 10 May 2017
Posts: 26

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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).

Intern  B
Joined: 17 Mar 2017
Posts: 9
GMAT 1: 510 Q47 V15 GMAT 2: 600 Q49 V22 ### Show Tags

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

?
Manager  G
Joined: 27 Dec 2016
Posts: 224
Concentration: Marketing, Social Entrepreneurship
GPA: 3.65
WE: Marketing (Education)

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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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Joined: 08 Jun 2015
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+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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Intern  B
Joined: 06 Nov 2016
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Bunuel Is 4C2 +3 correct for favorable outcomes?
Manager  G
Joined: 25 Apr 2018
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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
Veritas Prep GMAT Instructor V
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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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Location: United States (CA)

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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

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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Manager  G
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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... Bunuel
is this approach correct?
Math Expert V
Joined: 02 Sep 2009
Posts: 59634

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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... Bunuel
is this approach correct?

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Yes, it is.
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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. $$\frac{19}{20}$$ Re: M20-29   [#permalink] 26 Nov 2019, 00:57
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# M20-29

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