GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 10 Dec 2019, 19:36

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

M20-29

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59634

Show Tags

16 Sep 2014, 01:09
1
7
00:00

Difficulty:

25% (medium)

Question Stats:

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

HideShow timer Statistics

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. $$\frac{19}{20}$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 59634

Show Tags

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

_________________
Manager
Joined: 26 Apr 2012
Posts: 81
Location: India
Concentration: Entrepreneurship, General Management
GMAT 1: 640 Q48 V29
GMAT 2: 660 Q45 V35
GMAT 3: 680 Q48 V35
GPA: 2.8
WE: Information Technology (Computer Software)

Show Tags

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.

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 59634

Show Tags

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.

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.
_________________
Manager
Joined: 18 Jun 2016
Posts: 86
Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 700 Q49 V36

Show Tags

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

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
Joined: 02 Sep 2009
Posts: 59634

Show Tags

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

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.
_________________
Retired Moderator
Joined: 26 Nov 2012
Posts: 554

Show Tags

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...
Intern
Joined: 10 May 2017
Posts: 26

Show Tags

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).

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

Show Tags

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

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

Show Tags

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.
_________________
There's an app for that - Steve Jobs.
Senior Manager
Joined: 08 Jun 2015
Posts: 418
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

Show Tags

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.
_________________
" The few , the fearless "
Intern
Joined: 06 Nov 2016
Posts: 1

Show Tags

08 Aug 2019, 18:56
Bunuel Is 4C2 +3 correct for favorable outcomes?
Manager
Joined: 25 Apr 2018
Posts: 94
Location: India
Concentration: Finance, Technology
GMAT 1: 600 Q40 V33

Show Tags

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.

Bunuel, chetan2u, GMATPrepNow, MartyTargetTestPrep, ScottTargetTestPrep, VeritasKarishma
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9866
Location: Pune, India

Show Tags

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/
_________________
Karishma
Veritas Prep GMAT Instructor

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8659
Location: United States (CA)

Show Tags

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.

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.
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager
Joined: 25 Apr 2018
Posts: 94
Location: India
Concentration: Finance, Technology
GMAT 1: 600 Q40 V33

Show Tags

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...

Bunuel
is this approach correct?
Math Expert
Joined: 02 Sep 2009
Posts: 59634

Show Tags

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...

Bunuel
is this approach correct?

__________________
Yes, it is.
_________________
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5469
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Show Tags

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. $$\frac{19}{20}$$
Re: M20-29   [#permalink] 26 Nov 2019, 00:57
Display posts from previous: Sort by

M20-29

Moderators: chetan2u, Bunuel