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

 It is currently 16 Feb 2019, 21:48

### 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 Algebra Webinar

February 17, 2019

February 17, 2019

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Free GMAT Strategy Webinar

February 16, 2019

February 16, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# When 2 people are selected at random from a group of 4 females and 4 m

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6949
GMAT 1: 760 Q51 V42
GPA: 3.82
When 2 people are selected at random from a group of 4 females and 4 m  [#permalink]

### Show Tags

27 Sep 2018, 23:42
00:00

Difficulty:

15% (low)

Question Stats:

76% (01:08) correct 24% (02:08) wrong based on 52 sessions

### HideShow timer Statistics

[Math Revolution GMAT math practice question]

When $$2$$ people are selected at random from a group of $$4$$ females and $$4$$ males, what is the probability that at least one female is selected?

$$A. \frac{5}{14}$$
$$B. \frac{7}{14}$$
$$C. \frac{9}{14}$$
$$D. \frac{11}{14}$$
$$E. \frac{13}{14}$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Director Joined: 19 Oct 2013 Posts: 508 Location: Kuwait GPA: 3.2 WE: Engineering (Real Estate) When 2 people are selected at random from a group of 4 females and 4 m [#permalink] ### Show Tags 27 Sep 2018, 23:51 Total possibilities 8c2 = (8*7)/2 = 28 Now since we are interested in probability at least one is female We can find the probability that all 2 are males 4c2 = (4*3)/2 = 6 Now 6/28 = 3/14 Probability of at least one female = 1 - (3/14) = 11/14 Answer choice D Posted from my mobile device CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2775 Location: India GMAT: INSIGHT Schools: Darden '21 WE: Education (Education) Re: When 2 people are selected at random from a group of 4 females and 4 m [#permalink] ### Show Tags 27 Sep 2018, 23:54 MathRevolution wrote: [Math Revolution GMAT math practice question] When $$2$$ people are selected at random from a group of $$4$$ females and $$4$$ males, what is the probability that at least one female is selected? $$A. \frac{5}{14}$$ $$B. \frac{7}{14}$$ $$C. \frac{9}{14}$$ $$D. \frac{11}{14}$$ $$E. \frac{13}{14}$$ Probability of 1 Female in selected team = 1- Probability of NO Female in selected team Probability of NO Female in selected team = 4C2/8C2 = 6/28 = 3/14 i.e. Probability of Atleast 1 female = 1-(3/14) = 11/14 Answer: Option D ALTERNATIVE Probability of atleast 1 female in team = Probability of exactly 1 female + probability of exactly two female = [(4C1*4C1)+(4C2)] / 8C2 = (16+6)/28 = 22/28 = 11/14 Answer: Option D _________________ 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6949 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: When 2 people are selected at random from a group of 4 females and 4 m [#permalink] ### Show Tags 30 Sep 2018, 21:58 => The probability that at least one female is selected is equal to 1 minus the probability that two males are selected. There are 4C2 ways of selecting 2 males from the four males and 8C2 ways of selecting 2 people from the group of 8. Therefore, the probability of selecting two males from the group is 4C2/8C2, and the probability of selecting at least one female is 1 – 4C2/8C2$$= 1 – { \frac{(4*3)}{(1*2) } / { (8*7)/(1*2)} } = 1 – \frac{(4*3)}{(8*7)} = 1 – \frac{3}{14} = \frac{11}{14}$$ Therefore, the answer is D. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Re: When 2 people are selected at random from a group of 4 females and 4 m   [#permalink] 30 Sep 2018, 21:58
Display posts from previous: Sort by