BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Probability Question

Expert replies
by mindruna » Fri Aug 10, 2007 5:14 am
I still struggle with this type of question. Can anyone help? Thanks.


Two canoe riders must be selected from each of two groups of campers. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?

A) 1/6
B) 1/4
C) 2/7
D) 1/3
E) 1/2
Join the discussion
Source: — Problem Solving |

Re: Probability Question

by ratindasgupta » Fri Aug 10, 2007 6:19 am
mindruna wrote:I still struggle with this type of question. Can anyone help? Thanks.


Two canoe riders must be selected from each of two groups of campers. One group consists of three men and one woman, and the other group consists of two women and one man. What is the probability that two men and two women will be selected?

A) 1/6
B) 1/4
C) 2/7
D) 1/3
E) 1/2
I think the answer is E) 1/2

If 2M and 2W have to be selected keeping in mind 2 from each group then these are the following options
a) 2M from 1st group and 2W from 2nd group or
b) 1M & 1W from 1st group and 1M & 1W from the 2nd group or
c) 1W & 1M from the 1st group and 1W & 1M from the 2nd group or
d) 1M & 1W from 1st group and 1W & 1M from the 2nd group or
e) 1W & 1M from the 1st group and 1M & 1W from the 2nd group.
Whew!

so
a) 1/6
b) 1/12
c) 1/12
d) 1/12
e) 1/12

a + b + c + d + e = 1/2.

Wot's the OA?
Join the discussion

by mindruna » Fri Aug 10, 2007 6:31 am
Thanks Ratindasgupta!

1/2 is the OA. How are you getting 1/6 for a and 1/12 for the rest?
Join the discussion

by ratindasgupta » Fri Aug 10, 2007 6:50 am
mindruna wrote:Thanks Ratindasgupta!

1/2 is the OA. How are you getting 1/6 for a and 1/12 for the rest?
i was trying to avoid the gory details. hee hee. anyway here goes.

basically its a dependant event problem

if u need 2m from 1st group then its 3/4 x 2/3 ((Probability of selecting 1st man is 3/4 and 2nd man is 2/3)

and 2w from the 2nd group is 2/3 x 1/2 (Probability of selecting 1st woman is 2/3 and 2nd woman is 1/2)

so 3/4 x 2/3 x 2/3 x 1/2 = 1/6

and so on and so forth. Just list the probabilities down.

1st Group 2nd Group
a) 2M 2W 3/4 x 2/3 x 2/3 x 1/2 = 1/6
b) 1M & 1W 1M & 1W 3/4 x 1/3 x 1/3 x 1 = 1/12
c) 1W & 1M 1W & 1M 1/4 x 1 x 2/3 x 1/2 = 1/12
d) 1M & 1W 1W & 1M 3/4 x 1/3 x 2/3 / 1/2 = 1/12
e) 1W & 1M 1M & 1W 1/4 x 1 x 1/3 x 1 = 1/12

Once this is done, add all figures.

I'm sorry i dont have an easier way to solve this. I doubt there is one. Where is this problem from btw?
Join the discussion

by mindruna » Fri Aug 10, 2007 6:55 am
That's completely clear--thanks. I found the problem online while trying to improve my weak points. I really appreciate the help!
Join the discussion

by ratindasgupta » Fri Aug 10, 2007 6:58 am
mindruna wrote:That's completely clear--thanks. I found the problem online while trying to improve my weak points. I really appreciate the help!
hey no sweat! that's wot's this forum is for!
Join the discussion