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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

probability of red and ace ???

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Tue Sep 25, 2012 9:38 pm
When replaced:
probability of the 1st card to be red = 1/2
when we replaced, we will have the same set of cards
probability for an ace = 4/52
probability of both events to happen = 1/2 * 4/52 = 1/26

when not replaced:
probability of the 1st card to be red = 26/52
but the 1st card can be red ace too..
so we will divide it into two red cases..
1st one red suit without an ace, and second red ace
the probability of 1st card to be red non-ace = 24/52
the probability of 2nd card to be ace = 4/51
the probability that the 1st card is non-ace red card and second card is ace = 24/52 * 4/51
the probability of the 1st card to be red ace = 2/52
the probability of the 2nd card to be ace = 3/51
the probability that the 1st card is ace red card and second card is ace = 2/52 * 3/51

the probability that 1st card is red and the second is an Ace = the probability that the 1st card is non-ace red card and second card is ace + the probability that the 1st card is ace red card and second card is ace
= (24/52 * 4/51) + (2/52 * 3/51)
= 1/26
Last edited by Anurag@Gurome on Wed Sep 26, 2012 2:19 am, edited 1 time in total.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 26
Joined: Wed Jun 20, 2012 10:42 pm

by smanstar » Tue Sep 25, 2012 10:08 pm
Anurag@Gurome wrote:When replaced:
probability of the 1st card to be red = 1/4
when we replaced, we will have the same set of cards
probability for an ace = 4/52
probability of both events to happen = 1/4 * 4/52 = 1/52

when not replaced:
probability of the 1st card to be red = 13/52
but the 1st card can be red ace too..
so we will divide it into two red cases..
1st one red suit without an ace, and second red ace
the probability of 1st card to be red non-ace = 12/52
the probability of 2nd card to be ace = 4/51
the probability that the 1st card is non-ace red card and second card is ace = 12/52 * 4/51
the probability of the 1st card to be red ace = 1/52
the probability of the 2nd card to be ace = 3/51
the probability that the 1st card is ace red card and second card is ace = 1/52 * 3/51

the probability that 1st card is red and the second is an Ace = the probability that the 1st card is non-ace red card and second card is ace + the probability that the 1st card is ace red card and second card is ace
= 12/52 * 4/51 + 1/52 * 3/51
= 1/52



Sir

Can you explain how did you find the probability of red ?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Mon Jun 25, 2012 10:56 pmFollowed by: 10 members
Thanked: 60 times

by anuprajan5 » Tue Sep 25, 2012 10:21 pm
Anurag@Gurome wrote:When replaced:
probability of the 1st card to be red = 1/4

when not replaced:
probability of the 1st card to be red = 13/52
Anurag,

I am a bit stumped with this one. There are 26 red and black cards each in a deck. Therefore, probability of the 1st card to be red should be 26/52 = 1/2

Regards
Anup
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Wed Sep 26, 2012 2:14 am
Anurag,

I am a bit stumped with this one. There are 26 red and black cards each in a deck. Therefore, probability of the 1st card to be red should be 26/52 = 1/2

Regards
Anup
Its been long time i played cards. i was thinking of Red diamonds. Yeah the probability for a red is 1/2.
i will edit the ans.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1