|
Author |
Message |
|
TAGS:
|
|
|
Current Student
Joined: 31 Aug 2007
Posts: 374
Followers: 1
Kudos [?]:
36
[0], given: 1
|
Five marbles are in a bag: two are red and three are blue. [#permalink]
 03 Dec 2007, 13:12
Question Stats:
33% (01:30) correct
66% (00:22) wrong based on 0 sessions
Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.
1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20
|
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3596
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 231
Kudos [?]:
1300
[0], given: 346
|
C
1-3C2/5C2 = 1-3/10=7/10
3C2 - 2 blue of three
5C2 - the total number of combination
|
|
|
|
|
|
Current Student
Joined: 31 Aug 2007
Posts: 374
Followers: 1
Kudos [?]:
36
[0], given: 1
|
walker wrote: C
1-3C2/5C2 = 1-3/10=7/10
3C2 - 2 blue of tree
5C2 - total number of combination
agreed, supposedly the OA is 2/5...must be wrong.
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3596
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 231
Kudos [?]:
1300
[0], given: 346
|
young_gun wrote: agreed, supposedly the OA is 2/5...must be wrong.
hm... other way:
1-st marble:
1. [Red;2/5]
2. [Blue;3/5]
2-nd marble:
1. [Red;2/5][Red or Blue;1] = 2/5
2. [Blue;3/5][Red;2/4] = 6/20=3/10
p=2/5+3/10=7/10
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2530
Followers: 41
Kudos [?]:
358
[0], given: 19
|
Re: PS probability [#permalink]
03 Dec 2007, 16:26
young_gun wrote: Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.
1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20
at least 1 is red = 1 - no red
at least 1 is red = 1 - all blue
at least 1 is red = 1 - 3c2/5c2
at least 1 is red = 1 - 3/10 = 7/10
|
|
|
|
|
|
CEO
Joined: 29 Mar 2007
Posts: 2618
Followers: 13
Kudos [?]:
142
[0], given: 0
|
Re: PS probability [#permalink]
03 Dec 2007, 23:04
young_gun wrote: Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.
1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20
How bout none will be red. Very fast calculation then.
3/5*2/4 = 6/20 --> 14/20 --> 7/10. C
|
|
|
|
|
|
Director
Joined: 09 Aug 2006
Posts: 776
Followers: 1
Kudos [?]:
18
[0], given: 0
|
I suck at these but I did the following and got the same answer as OA:
You can either pick both red OR you can pick 1 red and 1 blue.
p(1st is red) * p(2nd is red) = 2/5 * 1/4 = 2/20
p(1st is red) * p(2nd is blue) = 2/5 * 3/4 = 6/20
Total p = 2/20 + 6/20 = 2/5
What's the source?
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3596
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 231
Kudos [?]:
1300
[0], given: 346
|
GK_Gmat wrote: I suck at these but I did the following and got the same answer as OA:
You can either pick both red OR you can pick 1 red and 1 blue.
p(1st is red) * p(2nd is red) = 2/5 * 1/4 = 2/20
p(1st is red) * p(2nd is blue) = 2/5 * 3/4 = 6/20
Total p = 2/20 + 6/20 = 2/5
What's the source?
You should add p(1st is blue) * p(2nd is red) = 3/5*2/4=6/20
2/5+6/20=7/10
|
|
|
|
|
|
Director
Joined: 09 Aug 2006
Posts: 776
Followers: 1
Kudos [?]:
18
[0], given: 0
|
walker wrote: GK_Gmat wrote: I suck at these but I did the following and got the same answer as OA:
You can either pick both red OR you can pick 1 red and 1 blue.
p(1st is red) * p(2nd is red) = 2/5 * 1/4 = 2/20
p(1st is red) * p(2nd is blue) = 2/5 * 3/4 = 6/20
Total p = 2/20 + 6/20 = 2/5
What's the source? You should add p(1st is blue) * p(2nd is red) = 3/5*2/4=6/20 2/5+6/20=7/10 Ok; didn't consider that. Thanks a lot! Agree on 7/10.
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1842
Location: New York
Followers: 20
Kudos [?]:
291
[0], given: 5
|
Re: PS probability [#permalink]
25 Aug 2008, 13:40
young_gun wrote: Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.
1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20 p=1- (3C2/ 5C2)=7/10 or p=(2C2*3C0+2C1*3C1)/ 5C2= (6+1)/10 =7/10
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
Manager
Joined: 22 Jul 2008
Posts: 156
Followers: 1
Kudos [?]:
5
[0], given: 0
|
Re: PS probability [#permalink]
25 Aug 2008, 14:33
You cannot do 2 C 1 and 3 C 1 in this case because they are identical marbles. Therefore, to choose 1 Red or 1 blue marble, there is only 1 way from amongst the Red or Blue marbles. However, the method used is very simple.
1 R + 1 B--- 2/5 * 3/4 = 6/20
2 R -----2/5 * 1/4 = 2/20
Therefore, total Probability = 6/20 + 2/20 = 8/20 = 2/5
|
|
|
|
|
|
Intern
Joined: 23 May 2008
Posts: 49
Followers: 0
Kudos [?]:
1
[0], given: 0
|
Re: PS probability [#permalink]
25 Aug 2008, 16:33
Can someone give the general rules for solving probability questions please?
|
|
|
|
|
|
Manager
Joined: 27 Oct 2008
Posts: 188
Followers: 1
Kudos [?]:
42
[0], given: 3
|
Re: PS probability [#permalink]
27 Sep 2009, 21:28
Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.
1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20
Soln:
Probability that atleast one will be red is
= Probability that exactly one is red + probability that both are red
= (2C1 * 3C1/5C2) + 2C2/5C2
= 7/10
|
|
|
|
|
|
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1400
Followers: 8
Kudos [?]:
84
[0], given: 10
|
Re: PS probability [#permalink]
02 May 2011, 20:02
1- [ 3c2/ 5c2] = 7/10
_________________
Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!
|
|
|
|
|
|
Manager
Joined: 09 Aug 2010
Posts: 107
Followers: 1
Kudos [?]:
11
[0], given: 7
|
Re: PS probability [#permalink]
02 May 2011, 21:29
Another approach:
R R B B B
Using Anagram Method:
How many ways can we select 2 of any color from 5? 5!/(2!3!) = 10
How many ways can we select (R,B) from R R B B B? 2!/1!1! x 3!/1!2! = 2 x 3 = 6
How many ways can we select (R,R) from R R B B B? 2!/2! = 1
Probability(of at least one Red) = (6 + 1)/10 = 7/10 OR C
|
|
|
|
|
|
|
Re: PS probability
[#permalink]
02 May 2011, 21:29
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
There are five balls in a bag. Two of them are red and three
|
Konstantin Lynov |
2 |
11 Aug 2003, 07:46 |
|
|
|
|
A bag contains six marbles: two red, two blue, and two
|
vivek_dj |
3 |
05 Jan 2004, 13:39 |
|
|
|
|
Bag A contains red, white and blue marbles such that the red
|
nxrfelix |
5 |
28 May 2007, 20:12 |
|
|
|
|
A jar contains two red marbles, three blue marbles, and four
|
snowbirdskier |
2 |
08 Sep 2008, 14:57 |
 |
6
|
 |
Bag A contains red, white and blue marbles such that the red
|
dred |
45 |
29 Jun 2007, 15:38 |
|
|
|
|
|
|
close
