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

 It is currently 22 Feb 2019, 17:43

### 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 February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
• ### FREE Quant Workshop by e-GMAT!

February 24, 2019

February 24, 2019

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

# Two hard candies are to be selected at random and without replacement

Author Message
TAGS:

### Hide Tags

Manager
Joined: 16 Aug 2014
Posts: 53
Two hard candies are to be selected at random and without replacement  [#permalink]

### Show Tags

15 Oct 2018, 03:57
1
00:00

Difficulty:

5% (low)

Question Stats:

91% (01:06) correct 9% (02:19) wrong based on 42 sessions

### HideShow timer Statistics

Two hard candies are to be selected at random and without replacement from a bag containing 5 green, 6 red, and 9 yellow candies. What is the probability that both candies selected will be red?

A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38

_________________

Hit Kudos if you like the post

Director
Joined: 19 Oct 2013
Posts: 508
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: Two hard candies are to be selected at random and without replacement  [#permalink]

### Show Tags

15 Oct 2018, 04:13
mayursurya wrote:
Two hard candies are to be selected at random and without replacement from a bag containing 5 green, 6 red, and 9 yellow candies. What is the probability that both candies selected will be red?

A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38

Total possible selection = 5+6+9 = 20

Total possible red = 6

The question asks for 2 red and it says without replacement so it will be in the form of $$\frac{R}{Total} * \frac{(R -1)}{Total}$$

Probability of selecting 2 red = $$\frac{6}{20} * \frac{5}{19}$$ = $$\frac{6}{4} * \frac{1}{19}$$ = $$\frac{3}{2} * \frac{1}{19}$$ = $$\frac{3}{38}$$

CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2799
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: Two hard candies are to be selected at random and without replacement  [#permalink]

### Show Tags

15 Oct 2018, 04:19
mayursurya wrote:
Two hard candies are to be selected at random and without replacement from a bag containing 5 green, 6 red, and 9 yellow candies. What is the probability that both candies selected will be red?

A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38

Total candies = 5+6+9 = 20

Red candies = 6

Favourable Outcomes = Total ways of selecting two red candies (out of 6) = 6C2 = 15

Total Outcomes = Total ways of selecting two candies (out of total 20 candies) = 20C2 = 190

Probability = favourable Outcomes / Total Outcomes = 15/190 = 3/38

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Manager
Joined: 24 Dec 2017
Posts: 182
Location: India
Concentration: Strategy, Real Estate
Schools: Johnson '21
Re: Two hard candies are to be selected at random and without replacement  [#permalink]

### Show Tags

15 Oct 2018, 07:39
Total candies = 20
Green = 5
Red = 6
Yellow = 9

Probability = Favourable/Total

=>Probability of selecting the first red candy = 6/20
=>Probability of selecting the second red candy = 5/19
=>6/20 x 5/19
=>30/380
=>3/38

Re: Two hard candies are to be selected at random and without replacement   [#permalink] 15 Oct 2018, 07:39
Display posts from previous: Sort by