It is currently 21 Oct 2017, 04:02

### 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

# prob - exactly vs. at least

Author Message
CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1049 [0], given: 4

Location: New York City
prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 08:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Please correct me if i am wrong.

A bag of 10 marbles contains 3 red marbles and 7 blue marbles.
If two marbles are selected at random, what is the probability that
a) exactly one is blue?
b) at least one marble is blue?

desired / total
a) 3C1 * 7C1 / 10C2 = 21/45

b) Total - P(none) / total
10C2 - 3C2 / 10C2 = 45-3/45 = 42/45 = 14/15
_________________

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

Kudos [?]: 1049 [0], given: 4

Director
Joined: 12 Jul 2007
Posts: 857

Kudos [?]: 330 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 08:39
looks good to me

Kudos [?]: 330 [0], given: 0

Intern
Joined: 13 Dec 2007
Posts: 28

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

Re: prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 09:04
a.
b. Probability that at least one marble is blue = 1 - (probability that none of the marble is blue)
= 1 - 7C2/10C2
= 1 - 21/45
= 24/45

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

Director
Joined: 12 Jul 2007
Posts: 857

Kudos [?]: 330 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 09:17
bmwhype2 wrote:
akhi wrote:
a.
b. Probability that at least one marble is blue = 1 - (probability that none of the marble is blue)
= 1 - 7C2/10C2
= 1 - 21/45
= 24/45

u selected 2 blue.

and subtracted incorrectly

Kudos [?]: 330 [0], given: 0

Intern
Joined: 13 Dec 2007
Posts: 28

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

Re: prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 09:21
eschn3am wrote:
bmwhype2 wrote:
akhi wrote:
a.
b. Probability that at least one marble is blue = 1 - (probability that none of the marble is blue)
= 1 - 7C2/10C2
= 1 - 21/45
= 24/45

u selected 2 blue.

and subtracted incorrectly

sorry..my bad..this is wht happens when u try to solve problem at work..
it should be
1-3C2 / 10C2 = 42/45

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

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4587 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 09:23
other ways:

a)
1. p=7/10*3/9+3/10*7/9=7*3*2/(10*9)=7/15
2. p=3P1*7P1*2P2/10P2=3*7*2/(10*9)=7/15

b)
1. p=1-3/10*2/9=1-1/15=14/15
2. p=1-3P2/10P2=1-3*2/(10*9)=14/15
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4587 [0], given: 360

SVP
Joined: 28 Dec 2005
Posts: 1545

Kudos [?]: 179 [0], given: 2

Re: prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 19:00
isnt the probability that at least one is blue equal to the sum of the probability that you have one blue, one red and the probability that both are blue ?

Kudos [?]: 179 [0], given: 2

Director
Joined: 12 Jul 2007
Posts: 857

Kudos [?]: 330 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

08 Jan 2008, 19:35
pmenon wrote:
isnt the probability that at least one is blue equal to the sum of the probability that you have one blue, one red and the probability that both are blue ?

Yes, but that's more work than finding finding out the probability of 2 red and just subtracting that from 1.

Kudos [?]: 330 [0], given: 0

Director
Joined: 09 Aug 2006
Posts: 754

Kudos [?]: 255 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 02:12
bmwhype2 wrote:
Please correct me if i am wrong.

A bag of 10 marbles contains 3 red marbles and 7 blue marbles.
If two marbles are selected at random, what is the probability that
a) exactly one is blue?
b) at least one marble is blue?

desired / total
a) 3C1 * 7C1 / 10C2 = 21/45

Getting 7/15 for a.

p(exactly 1 blue) = 7/10 * (1-6/9) * 2! = 7/15

Kudos [?]: 255 [0], given: 0

Director
Joined: 09 Jul 2005
Posts: 589

Kudos [?]: 64 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 03:45
P(one is blue)=(3/10)*(7/10)+(7/10)*(3/10)=21/50*2=7/15
P(at least one is blue)=1-P(red&red)=1-(3/10)*(2/9)=14/15

Or still another way: Probability=#favourable/#total

P(one is blue)=[2!*C(7,1)*C(3,1)]/[10!/8!]=21/45=7/15

Last edited by automan on 09 Jan 2008, 04:05, edited 1 time in total.

Kudos [?]: 64 [0], given: 0

VP
Joined: 22 Nov 2007
Posts: 1079

Kudos [?]: 679 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 03:56
automan wrote:
P(one is blue)=(3/10)*(7/10)+(7/10)*(3/10)=21/50*2=7/15
P(at least one is blue)=1-P(red&red)=1-(3/10)*(2/9)=14/15

yes, walker is correct

Kudos [?]: 679 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1049 [0], given: 4

Location: New York City
Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 06:28
GK_Gmat wrote:
bmwhype2 wrote:
Please correct me if i am wrong.

A bag of 10 marbles contains 3 red marbles and 7 blue marbles.
If two marbles are selected at random, what is the probability that
a) exactly one is blue?
b) at least one marble is blue?

desired / total
a) 3C1 * 7C1 / 10C2 = 21/45

Getting 7/15 for a.

p(exactly 1 blue) = 7/10 * (1-6/9) * 2! = 7/15

thats the same value. 21/45 = 7/15
_________________

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

Kudos [?]: 1049 [0], given: 4

CEO
Joined: 29 Mar 2007
Posts: 2554

Kudos [?]: 516 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 08:19
bmwhype2 wrote:
Please correct me if i am wrong.

A bag of 10 marbles contains 3 red marbles and 7 blue marbles.
If two marbles are selected at random, what is the probability that
a) exactly one is blue?
b) at least one marble is blue?

desired / total
a) 3C1 * 7C1 / 10C2 = 21/45

b) Total - P(none) / total
10C2 - 3C2 / 10C2 = 45-3/45 = 42/45 = 14/15

A: consider two possiblities: 3/10*7/9 (2) --> 21/45 --> 7/15 b/c we have to consider the scenario of 7/10*3/9. So the multiplied by 2 is the same as 21/90+21/90 --> 42/90 --> 7/15.

B: 1-prob. that none will be blue. 3/10*2/9 --> 6/90 --> 15/15-1/15 --> 14/15

Kudos [?]: 516 [0], given: 0

Manager
Joined: 04 Jan 2008
Posts: 83

Kudos [?]: 14 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 13:55
Walker is right
for qn1. for exactly one to be blue, it should be BR & RB 7/10*3/8 + 3/10*7/9 = 7/15
for qn2. atleast one Blue, you can add BB to the above & get the answer.
prob of BB is 7/10*6/9 =7/15
so total of BB< RB and BR = 2*7/15= 14/15

Kudos [?]: 14 [0], given: 0

Manager
Joined: 26 Dec 2007
Posts: 70

Kudos [?]: 25 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 18:31
Sorry this might just be my ingnorance, but what does "p1... p2" stand for??

a)
1. p=7/10*3/9+3/10*7/9=7*3*2/(10*9)=7/15
2. p=3P1*7P1*2P2/10P2=3*7*2/(10*9)=7/15

b)
1. p=1-3/10*2/9=1-1/15=14/15
2. p=1-3P2/10P2=1-3*2/(10*9)=14/15

Kudos [?]: 25 [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4587 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: prob - exactly vs. at least [#permalink]

### Show Tags

09 Jan 2008, 23:51
bsjames2 wrote:
Sorry this might just be my ingnorance, but what does "p1... p2" stand for??

a)
1. p=7/10*3/9+3/10*7/9=7*3*2/(10*9)=7/15
2. p=3P1*7P1*2P2/10P2=3*7*2/(10*9)=7/15

b)
1. p=1-3/10*2/9=1-1/15=14/15
2. p=1-3P2/10P2=1-3*2/(10*9)=14/15

3P1 - permutation formula: 3!/2!
nPm=n!/(n-m)!
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4587 [0], given: 360

Senior Manager
Joined: 07 Jan 2008
Posts: 286

Kudos [?]: 48 [0], given: 0

Re: prob - exactly vs. at least [#permalink]

### Show Tags

10 Jan 2008, 12:02
1. RB + BR = 7/10 * 3/9 + 3/10 * 7/9 = 7/15

2. RB + BR + RR = 7/15 (from above) + 7/10 * 6/9
= 7/15 + 14/30 = 28/30 = 14/15

OR
1 - RR [All cases except when both picks are red]
1 - 3/10 * 2/9 = 1 - 1/15 = 14/15

Kudos [?]: 48 [0], given: 0

Re: prob - exactly vs. at least   [#permalink] 10 Jan 2008, 12:02
Display posts from previous: Sort by