Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart 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

PROBABILITY DS

Expert replies
by cramya » Sat Nov 15, 2008 2:03 pm
If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?

1) There are 41 male students in the class.
2) The probability of selecting one male and one female student is 15/41.

[spoiler]OA: B[/spoiler]
Join the discussion
Source: — Data Sufficiency |

Re: PROBABILITY DS

by logitech » Sat Nov 15, 2008 2:44 pm
cramya wrote:If two students are chosen at random with replacement from a certain class, what is the probability that two male students or two female students are selected?

1) There are 41 male students in the class.
2) The probability of selecting one male and one female student is 15/41.

[spoiler]OA: B[/spoiler]
Cramya in da house!!

Okay let me try this:

M=Males
F=Females

What we need is:

P(2males) + P(2females)

With replacement means there will always be M+F students ?

=[M/(M+F) x M/(M+F)] + [F/(M+F) x F/(M+F)]

= (M^2+F^2)/[(M+F)^2]

Statement 1)

M=41 - INSUF

Statement 2)

The probability of selecting one male and one female student is 15/41.

=[M/(M+F) x F/(M+F)] = 15/41

=MF/[(M+F)^2] = 15/41

So, MF = 15X

[(M+F)^2] = 41X

Lets go back to our question stem:

(M^2+F^2)/[(M+F)^2]

= (M^2+F^2) = [(M+F)^2] - 2MF = 41X - 2(15X) = 11X

So;

(M^2+F^2)/[(M+F)^2] = 11X/41X = 11/41

SUFF

Hence, B

Does my solution make sense ?
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Join the discussion

by cramya » Sat Nov 15, 2008 2:54 pm
B is right but the I remember seeing the prob value to be 1-15/41 = 26/41

I will check but good approach.
The probability of selecting one male and one female student is 15/41.

=[M/(M+F) x F/(M+F)] = 15/41

=MF/[(M+F)^2] = 15/41
I am thinking this will be

[M/(M+F) x F/(M+F)] + [F/(M+F) x M/(M+F)] = 15/41
(Pick a male first then female or female first and then male)

It might get a bit complicated with this...Thoughts welcome.
Last edited by cramya on Sat Nov 15, 2008 3:12 pm, edited 3 times in total.
Join the discussion

by logitech » Sat Nov 15, 2008 3:05 pm
cramya wrote:B is right but the I remember seeing the prob value to be 1-15/41 = 26/41

I will check but good approach.
That's absolutely true! I am trying to find where I made a mistake.

Three ways: MM, FF, MF

p(MM) + p(FF) +p(MF) = 1

Perfect GMAT solution by the way.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Join the discussion

by cramya » Sat Nov 15, 2008 3:07 pm
That's absolutely true! I am trying to find where I made a mistake.

Three ways: MM, FF, FM

I think u missed FM i.e

Sample space

Four ways: MM, FF, MF,FM
Last edited by cramya on Sat Nov 15, 2008 3:25 pm, edited 1 time in total.
Join the discussion

by logitech » Sat Nov 15, 2008 3:20 pm
cramya wrote:
That's absolutely true! I am trying to find where I made a mistake.

Three ways: MM, FF, MF

I think u missed MM i.e

Sample space

Four ways: MM, FF, MF,MM
I don't follow you. We have two MM's and only one FF ?
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Join the discussion

by cramya » Sat Nov 15, 2008 3:27 pm
I don't follow you. We have two MM's and only one FF ?
Oops typo with all these M's and F's. Corrected it.
Join the discussion