|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 10 Dec 2007
Posts: 52
Followers: 0
Kudos [?]:
5
[1] , given: 0
|
2 integers will be randomly selected from the sets above [#permalink]
 17 Jan 2008, 14:50
1
This post received
KUDOS
Question Stats:
88% (01:51) correct
11% (01:30) wrong based on 1 sessions
A = {2,3,4,5} B = {4,5,6,7,8} 2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33
|
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3591
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 230
Kudos [?]:
1298
[0], given: 346
|
Re: PS - probability [#permalink]
17 Jan 2008, 14:59
Bp=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20or p=\frac{4}{4}*\frac{1}{5}=0.20or p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20
_________________
iPhone/iPod/iPad: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
Android: GMAT ToolKit (NEW!). POLL: What tool do you need next?
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Director
Joined: 01 Jan 2008
Posts: 635
Followers: 3
Kudos [?]:
114
[2] , given: 1
|
Re: PS - probability [#permalink]
17 Jan 2008, 15:26
2
This post received
KUDOS
You don't need to perform any calculations in this example. Regardless of the number you choose from set A, there is only one number (out of 5) that makes the sum equal 9. Therefore, the probability is 1/5. The answer is B.
|
|
|
|
|
|
Manager
Joined: 04 Jan 2008
Posts: 86
Followers: 1
Kudos [?]:
10
[0], given: 0
|
Re: PS - probability [#permalink]
22 Jan 2008, 17:10
B
total 20 ways to sum,
4 of the sums equals 9
so, prob is 4/20= 1/5= .20
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1841
Location: New York
Followers: 20
Kudos [?]:
289
[0], given: 5
|
Re: PS - probability [#permalink]
24 Aug 2008, 14:52
vermatanya wrote: Quote: A = {2,3,4,5}
B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15
B. 0.20
C. 0.25
D. 0.30
E. 0.33
-thanx
p= 4/(4c1*5c1)=1/5=0.2
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
Manager
Joined: 10 Jul 2009
Posts: 136
Location: Ukraine, Kyiv
Followers: 2
Kudos [?]:
14
[1] , given: 60
|
Re: PS - probability [#permalink]
07 Sep 2009, 11:29
1
This post received
KUDOS
Maybe I am late with response, but anyway. We have two sets with 4 and 5 integers included in each one respectively. Thus, we have 4*5=20 possible ways of selecting two integers from these sets. The requirement is that we need the sum of 9. Here we have 4 possible choices: 2 and 7 3 and 6 4 and 5 5 and 4 Probability = favourable / total 4/20 = 0.2
_________________
Never, never, never give up
|
|
|
|
|
|
Manager
Joined: 27 Oct 2008
Posts: 188
Followers: 1
Kudos [?]:
42
[0], given: 3
|
Re: PS - probability [#permalink]
27 Sep 2009, 06:17
A = {2,3,4,5}
B = {4,5,6,7,8}
Total number of possibilities of choosing two numbers is = 4 * 5 = 20 ways
No. of possibilities of choosing numbers such that their sum is 9 is:
(2,7),(3,6),(4,5),(5,4)
thus 4 ways
So probability is = 4/20 = 1/5 = 0.2
|
|
|
|
|
|
Senior Manager
Joined: 22 Dec 2009
Posts: 368
Followers: 9
Kudos [?]:
136
[0], given: 47
|
Re: PS - probability [#permalink]
14 Feb 2010, 13:24
vermatanya wrote: A = {2,3,4,5}
B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15
B. 0.20
C. 0.25
D. 0.30
E. 0.33
-thanx
Total ways of choosing 1 number from Set A and 1 from Set B.... 4c1 x 5c1 = 20 Comb which yield 9 as the sum - 2,7.....3,6.....4,5....5,4 = 4 comb... Probability = 4 / 20 = .20 = B
_________________
Cheers!
JT...........
If u like my post..... payback in Kudos!! 
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
~~Better Burn Out... Than Fade Away~~
|
|
|
|
|
|
Senior Manager
Joined: 01 Feb 2010
Posts: 275
Followers: 1
Kudos [?]:
30
[0], given: 2
|
Re: PS - probability [#permalink]
15 Feb 2010, 00:24
vermatanya wrote: Quote: A = {2,3,4,5}
B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?
A. 0.15
B. 0.20
C. 0.25
D. 0.30
E. 0.33
-thanx
P(A) = 1 P(B) = 1/5 P(A&B) = 1/5 = 0.2 hence B
|
|
|
|
|
|
Intern
Joined: 24 Feb 2012
Posts: 36
Followers: 0
Kudos [?]:
3
[0], given: 18
|
Re: PS - probability [#permalink]
24 Feb 2012, 21:54
walker wrote: B
p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20
or
p=\frac{4}{4}*\frac{1}{5}=0.20
or
p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20 Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please?
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11518
Followers: 1792
Kudos [?]:
9538
[3] , given: 826
|
Re: PS - probability [#permalink]
24 Feb 2012, 22:49
3
This post received
KUDOS
fortsill wrote: walker wrote: B
p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20
or
p=\frac{4}{4}*\frac{1}{5}=0.20
or
p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20 Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please? I'd offer another approach which I hope will also clarify the above. A = {2,3,4,5}
B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 Rearrange the first set: A = {5, 4, 3, 2}B = {4, 5, 6, 7, 8}As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2. Answer: B. Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5. Answer: B. Hope it's clear.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Manager
Joined: 08 Jun 2011
Posts: 98
Followers: 1
Kudos [?]:
11
[0], given: 65
|
Re: PS - probability [#permalink]
03 May 2012, 05:11
Bunuel wrote: fortsill wrote: walker wrote: B
p=\frac{C^4_1}{C^4_1*C^5_1}=\frac{4}{4*5}=0.20
or
p=\frac{4}{4}*\frac{1}{5}=0.20
or
p=\frac{4}{5}*\frac{1}{4}+\frac{1}{5}*\frac{0}{4}=0.20 Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please? I'd offer another approach which I hope will also clarify the above. A = {2,3,4,5}
B = {4,5,6,7,8}
2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9?A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 Rearrange the first set: A = {5, 4, 3, 2}B = {4, 5, 6, 7, 8}As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 4*5=20. P=favorable/total=4/20=0.2. Answer: B. Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5. Answer: B. Hope it's clear. Bunuel, I am confused. We are selecting a pair from a total of 9 numbers or 7 numbers (if we eliminate duplicates). So we will be getting for total number of ways, in either case 2 from 9 is 36 and 2 from 7 is 21 So we have [ (2, 7),(3,6), (4,5) ] 3 pairs or [ (2, 7),(3,6), (4,5) (5,4) ] 4 pairs if we don't eliminate duplicates Either way, I am getting the wrong answer 3/36 and 4/21 are both wrong. Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same. How is it different from this one (the general concept)? if-two-of-the-four-expressions-x-y-x-5y-x-y-5x-y-are-92727.html#p713823Thank you
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3105
Location: Pune, India
Followers: 567
Kudos [?]:
1997
[1] , given: 92
|
Re: PS - probability [#permalink]
03 May 2012, 09:29
1
This post received
KUDOS
Lstadt wrote:
2 from 9 is 36
and
2 from 7 is 21
You are not selecting 2 numbers from 9 numbers. You are selecting 1 number from 4 numbers and 1 number from 5 numbers. Mind you, they are different. While selecting 2 numbers from {1, 2, 3, 4, 5, 6, 7, 8, 9}, you can select (3, 4) While selecting 1 from 4 and 1 from 5: {1, 2, 3, 4} and {5, 6, 7, 8, 9}, you cannot select (3, 4). That's the problem number 1. Lstadt wrote: Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same.
Secondly, the two given sets are distinct: A = {2,3,4,5} B = {4,5,6,7,8} When we pick a number from A and from B, we get {4, 5} or we can do it in this way {5, 4}. These are two different cases. In the first case, the 4 comes from A and in the second case, it comes from B.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3105
Location: Pune, India
Followers: 567
Kudos [?]:
1997
[0], given: 92
|
Re: 2 integers will be randomly selected from the sets above [#permalink]
20 Aug 2012, 04:12
Responding to a pm: No, there will not be 8 different pairs. A = {2,3,4,5} B = {4,5,6,7,8} If you select 2 from A, from B you must select 7. If you select 3 from A, from B you must select 6. If you select 4 from A, from B you must select 5. If you select 5 from A, from B you must select 4. The total number of ways in which the sum can be 9 is 4. The question was whether the last two cases are distinct or not. Is 4 from A and 5 from B the same as 4 from B and 5 from A. Since 4 and 5 come from different sets in the two cases, they represent two different cases and should be counted as such.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Intern
Joined: 12 Oct 2012
Posts: 18
WE: General Management (Hospitality and Tourism)
Followers: 0
Kudos [?]:
5
[0], given: 38
|
Re: 2 integers will be randomly selected from the sets above [#permalink]
15 Nov 2012, 03:25
HI Karishma,
I understood the question, got 0.20, but I have a doubt.
At first, I got 6 pairs - (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options.
How do we come too know that we need not include duplicate pairs??
I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q?
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3105
Location: Pune, India
Followers: 567
Kudos [?]:
1997
[0], given: 92
|
Re: 2 integers will be randomly selected from the sets above [#permalink]
15 Nov 2012, 03:33
aditi2013 wrote: HI Karishma,
I understood the question, got 0.20, but I have a doubt.
At first, I got 6 pairs - (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options.
How do we come too know that we need not include duplicate pairs??
I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q? You need to select a pair {a, b} such that their sum is 9. a is selected from set A and b is selected from set B. a can take a value from set A only i.e. one of 2/3/4/5 b can take a value from set B only i.e. one of 4/5/6/7/8 How do you select {6, 3}? a cannot be 6. A selection is different from another when you select different numbers from each set. One selection is {4, 5} - 4 from A, 5 from B Another selection is {5, 4} - 5 from A, 4 from B One selection is {3, 6} - 3 from A, 6 from B There is no such selection {6, 3} - A has no 6.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Intern
Joined: 12 Oct 2012
Posts: 18
WE: General Management (Hospitality and Tourism)
Followers: 0
Kudos [?]:
5
[0], given: 38
|
Re: 2 integers will be randomly selected from the sets above [#permalink]
15 Nov 2012, 03:40
when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it
cannot be (b,a) and it will always be (a,b) in terms of order??
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3105
Location: Pune, India
Followers: 567
Kudos [?]:
1997
[0], given: 92
|
Re: 2 integers will be randomly selected from the sets above [#permalink]
15 Nov 2012, 03:46
aditi2013 wrote: when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it
cannot be (b,a) and it will always be (a,b) in terms of order?? It doesn't matter how you write it. One number is from A and one is from B. You know which number is from A and which is from B. You can write it is {a, b} or {b, a}, it doesn't matter. {4, 5} - 4 is from A and 5 is from B {5, 4} - 5 is from A and 4 is from B These two are different. {2, 6} - 2 is from A and 6 is from B {6, 2} - 2 is from A and 6 is from B These two are same.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews
|
|
|
|
|
|
Intern
Joined: 12 Oct 2012
Posts: 18
WE: General Management (Hospitality and Tourism)
Followers: 0
Kudos [?]:
5
[0], given: 38
|
Re: 2 integers will be randomly selected from the sets above [#permalink]
15 Nov 2012, 03:51
VeritasPrepKarishma wrote: aditi2013 wrote: when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it
cannot be (b,a) and it will always be (a,b) in terms of order?? It doesn't matter how you write it. One number is from A and one is from B. You know which number is from A and which is from B. You can write it is {a, b} or {b, a}, it doesn't matter. {4, 5} - 4 is from A and 5 is from B {5, 4} - 5 is from A and 4 is from B These two are different. {2, 6} - 2 is from A and 6 is from B {6, 2} - 2 is from A and 6 is from B These two are same. Thanks karishma. Hope i master these concepts soon.
|
|
|
|
|
|
|
Re: 2 integers will be randomly selected from the sets above
[#permalink]
15 Nov 2012, 03:51
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
If two different numbers are randomly selected from a set
|
BG |
9 |
02 Jun 2005, 09:08 |
|
|
|
|
If two numbers are randomly selected from set (1, 2, 3, 4,
|
bmwhype2 |
1 |
23 Oct 2007, 12:32 |
|
|
2
|
 |
If an integer is to be randomly selected from set M above
|
marcodonzelli |
9 |
12 Jan 2008, 05:29 |
|
|
2
|
|
Two integers will be randomly selected from sets A and B
|
2013gmat |
5 |
24 Oct 2012, 22:05 |
|
|
|
|
Two integers will be randomly selected from the sets above,
|
Walkabout |
2 |
02 Dec 2012, 07:51 |
|
|
|
|
|
|
close
