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

 It is currently 19 Sep 2018, 13:41

### 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

# Lily has only red and blue balls in a jar. If she removes three balls,

Author Message
TAGS:

### Hide Tags

VP
Joined: 23 Sep 2015
Posts: 1041
Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

### Show Tags

10 Jun 2018, 07:21
1
2
00:00

Difficulty:

95% (hard)

Question Stats:

30% (10:59) correct 70% (02:09) wrong based on 82 sessions

### HideShow timer Statistics

Lily has only red and blue balls in a jar. If she removes three balls, is the probability of getting all red balls greater than the probability of getting at least one blue ball?

(1) The number of red balls is more than three times the number of blue balls.
(2) Less than (1)/(4) th of the balls in the jar are blue.

Source: Expert Global

_________________

Thanks!
Do give some kudos.

Simple strategy:
“Once you’ve eliminated the impossible, whatever remains, however improbable, must be the truth.”

Best Gmat Resource:
GmatPrep CR|GmatPrep SC|GmatPrep RC

My Notes:

Manager
Joined: 19 Feb 2017
Posts: 133
Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

### Show Tags

10 Jun 2018, 08:50
Answer should be E. Eg: Take there are 4 red balls and 1 blue ball. Then P(RRR)= 2/5 and P(atleast 1B) = 3/5. So 2nd expression is smaller.

Then take there are 7 red balls and 1 blue ball. Then P(RRR)= 5/8 and P(atleast 1B) = 3/8. So 1st expression is smaller.

This example will work for both the statements. Therefore, no sufficiency in any. Hence, E.
Intern
Joined: 03 Sep 2015
Posts: 17
Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

### Show Tags

28 Aug 2018, 10:13
Hi. How did you get 3/5 for blue balls?
Intern
Joined: 09 Aug 2018
Posts: 3
Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

### Show Tags

29 Aug 2018, 05:37
Hi
I find the resource helpful, you can also try this probability lecture.
Intern
Joined: 03 Sep 2015
Posts: 17
Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

### Show Tags

29 Aug 2018, 05:48
Thanks so much for this

Posted from my mobile device
GMAT Tutor
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 168
Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

### Show Tags

29 Aug 2018, 06:57
1
Quote:
Lily has only red and blue balls in a jar. If she gets three balls randomly, is the probability of getting all red balls greater than the probability of getting at least one blue ball?

(1) The number of red balls is more than three times the number of blue balls.
(2) Less than (1)/(4) of the balls in the jar are blue.

Source: Expert Global

Nice problem!

$$\operatorname{P} \left( {{\text{all}}\,\,3\,\,{\text{red}}} \right)\,\,\,\,\mathop > \limits^? \,\,\,\operatorname{P} \left( {{\text{not}}\,\,{\text{all}}\,\,3\,\,{\text{red}}} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\operatorname{P} \left( {{\text{all}}\,\,3\,\,{\text{red}}} \right) = \boxed{\frac{{C\left( {r,3} \right)}}{{C\left( {r + b,3} \right)}}\,\,\,\,\mathop > \limits^? \,\,\,\frac{1}{2}}$$

$$\left( {1 + 2} \right)\,\,$$

$$Take\,\,\left( {r,b} \right) = \left( {4,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {4,3} \right)}}{{C\left( {5,3} \right)}} = \,\frac{1}{{10}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle$$

$$Take\,\,\left( {r,b} \right) = \left( {5,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {5,3} \right)}}{{C\left( {6,3} \right)}} = \,\frac{{10}}{{20}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,{\text{but}}\,\,\,{\text{increasing}}\,\,...$$

$$Take\,\,\left( {r,b} \right) = \left( {6,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {6,3} \right)}}{{C\left( {7,3} \right)}} = \,\frac{4}{7}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,$$

(Keeping the number of blue balls fixed and increasing the number of red ones, we know the probability of getting 3 reds must increase...)

The answer is (E), because we were able to BIFURCATE (1+2)!

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 30/Sep with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 70% discount!

Lily has only red and blue balls in a jar. If she removes three balls, &nbs [#permalink] 29 Aug 2018, 06:57
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.