Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 07 May 2015, 01:40

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Marbles

Author Message
TAGS:
Manager
Joined: 05 Nov 2005
Posts: 232
Location: Germany
Followers: 2

Kudos [?]: 18 [0], given: 0

Marbles [#permalink]  18 Dec 2005, 07:40
1
This post was
BOOKMARKED

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
Director
Joined: 17 Dec 2005
Posts: 552
Location: Germany
Followers: 1

Kudos [?]: 12 [0], given: 0

Same procedure as with PS 1-29 I think,

We must add the propability that,

one marble is red: 2/5 * 3/4= 6/20
with
two marbles are red: 2/5 * 1/4= 2/20

Which gives 2/5, so it is A.
Manager
Joined: 02 Jun 2005
Posts: 56
Followers: 1

Kudos [?]: 0 [0], given: 0

Re: Marbles [#permalink]  18 Dec 2005, 08:38
sandalphon 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

Vote for 3)

Andy pick 2 out of 5 without replacing the marbles back into the bag. The chances of picking up 2 blue marbles are 3/5*1/2=3/10, thus the probability pikcing up at least 1 red marble is 1-3/10=7/10.
Director
Joined: 17 Dec 2005
Posts: 552
Location: Germany
Followers: 1

Kudos [?]: 12 [0], given: 0

@tinyseal,

can you see why we have different results, while our approaches are directly reverse? You come from behind and I'm from the front. Both correct. Can't see the problem.
Manager
Joined: 02 Jun 2005
Posts: 56
Followers: 1

Kudos [?]: 0 [0], given: 0

two marbles are red: 2/5 * 1/4= 2/20

allabout, I think that the probility of picking up two red marbles is: 2/5.
Director
Joined: 17 Dec 2005
Posts: 552
Location: Germany
Followers: 1

Kudos [?]: 12 [0], given: 0

Intern
Joined: 05 Dec 2005
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 0

I think 2/5 ..as said by allabout

To elucidate

The event can be seen as

1.atleast 1 red marble + no red marble .

prob of drawing atleast 1 red marble + probablity of no red marble(meaning that the two marbles drawn are blue) = 1 ....(a)

probablity of drawing 2 blue marbles = 3/5
hence prob of atleast on red marble = 1-3/5 = 2/5

thanx
_________________

Any one can know the seeds in an apple ,BUT only GOD know the number of apples in the seed

Senior Manager
Joined: 14 Apr 2005
Posts: 419
Location: India, Chennai
Followers: 1

Kudos [?]: 6 [0], given: 0

Re: Marbles [#permalink]  19 Dec 2005, 00:44
sandalphon 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(Atleast one red) = 1- probablity (no red)
= 1 - 3/5
= 2/5
Intern
Joined: 16 Oct 2005
Posts: 27
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Marbles [#permalink]  19 Dec 2005, 14:14
krisrini wrote:
P(Atleast one red) = 1- probablity (no red)
= 1 - 3/5
= 2/5

Krisrini, it's a smart way of doing it..

I calculated this as
Probability that both are red + Probability that one is read = 2/5*1/4 + 2/5 * 3/4 = 2/20 + 6/20 = 2/5
Senior Manager
Joined: 05 Oct 2005
Posts: 485
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: Marbles [#permalink]  19 Dec 2005, 15:17
strange wrote:
krisrini wrote:
P(Atleast one red) = 1- probablity (no red)
= 1 - 3/5
= 2/5

Krisrini, it's a smart way of doing it..

I calculated this as
Probability that both are red + Probability that one is read = 2/5*1/4 + 2/5 * 3/4 = 2/20 + 6/20 = 2/5

It should be 7/10.

Probability of no reds is not 3/5. It is 3/5 *2/4 = 3/10
1 - Prob of red = 1 - 3/10 = 7/10

Different approach:
- There are 5 balls, thus there are 5C2 ways to pick 2, or (5*4)/2! = 10.
- There are 3C2 ways to pick only blues (3*2)/2! = 3.
- (Total combinations) - (combinations with no reds) = 10 - 3 = 7
- Therefore, 7/10 is the prob of picking at least 1 red.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
Followers: 22

Kudos [?]: 184 [0], given: 0

P(at least one is red) = 1-P(pick 2 blue)
P(pick 2 blue) = P(none is red)

P(pick 2 blue) = 3/5*2/4 = 3/10
P(at least 1 red) = 1-3/10 = 7/10
Senior Manager
Joined: 09 Aug 2005
Posts: 286
Followers: 1

Kudos [?]: 2 [0], given: 0

total combos = 5C2 = 10

good results

R1R2

R1B1 R1B2 R1B3

R2B1, R2B2, R2B3

7 in all

so 7/10

in GMAT usually numbers are small - I find this approach more confidance building and convincing.
SVP
Joined: 28 May 2005
Posts: 1743
Location: Dhaka
Followers: 6

Kudos [?]: 72 [0], given: 0

it can be 1R1B or 1B1R or 2R

= 2/5*1/4 + 3/5*2/4 + 2/5*3/4
= 14/20 or 7/10
_________________

hey ya......

Director
Joined: 17 Dec 2005
Posts: 552
Location: Germany
Followers: 1

Kudos [?]: 12 [0], given: 0

nakib77 wrote:
it can be 1R1B or 1B1R or 2R

= 2/5*1/4 + 3/5*2/4 + 2/5*3/4
= 14/20 or 7/10

That's the point why I think 2/5 is right.

the order in which we choose the two marbles doesn't matter, so that we have not to consider RB and BR as different possibilities.

Hope sandalphon will clear it up soon, I'm curious about the result.
Director
Joined: 09 Jul 2005
Posts: 595
Followers: 2

Kudos [?]: 26 [0], given: 0

P(at least 1 red)= 1 - P(no red marble) = 1- (3/5)*(2/4)=14/20=7/10

C.
Manager
Joined: 05 Nov 2005
Posts: 232
Location: Germany
Followers: 2

Kudos [?]: 18 [0], given: 0

OA is A, altought I have great problems deciphering that question. I have no OE since it is ffrom the Princeton CD, sorry guys.
CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 350 [0], given: 4

Re: [#permalink]  23 Feb 2008, 20:11
sandalphon wrote:
OA is A, altought I have great problems deciphering that question. I have no OE since it is ffrom the Princeton CD, sorry guys.

there is no way that the OA is A.

A is only possible if the question asked what is the probability of choosing 1 red marble out of the total.

at least one red = 1 - (3/5)(2/4)
OR
at least one red = (5C2 - 3C2)/(5C2)
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Manager
Joined: 20 Sep 2007
Posts: 107
Followers: 1

Kudos [?]: 33 [0], given: 0

Re: Marbles [#permalink]  23 Feb 2008, 20:19
Yeah there is something wrong with the question or the answer. For probability for atleast one red from two should be 7/10
Re: Marbles   [#permalink] 23 Feb 2008, 20:19
Similar topics Replies Last post
Similar
Topics:
3 The contents of a bag of marbles consist of 9 green marbles 10 16 Mar 2012, 03:24
There are 20 red marbles, 20 blue marbles, 5 white marbles, 5 22 Jul 2008, 06:39
5 blue marbles, 3 red marbles and 4 purple marbles are 7 25 Jan 2008, 09:44
Probability .. Marbles 5 23 Nov 2006, 03:49
18 5 blue marbles, 3 red marbles and 4 purple marbles are 19 16 Oct 2006, 10:42
Display posts from previous: Sort by

# Marbles

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.