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 →

Bill has a set of \(6\) black cards and a set of \(6\) red cards. Each card has a number from \(1\) through \(6,\) such

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2898
Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members
Thanked: 6 times
by Vincen » Thu May 13, 2021 6:09 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

Bill has a set of \(6\) black cards and a set of \(6\) red cards. Each card has a number from \(1\) through \(6,\) such that each of the numbers \(1\) through \(6\) appears on \(1\) black card and \(1\) red card. Bill likes to play a game in which he shuffles all \(12\) cards, turns over \(4\) cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

(A) \(\dfrac8{33}\)

(B) \(\dfrac{62}{65}\)

(C) \(\dfrac{17}{33}\)

(D) \(\dfrac{103}{165}\)

(E) \(\dfrac{25}{33\})

Answer: C

Source: Manhattan GMAT
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

Re: Bill has a set of \(6\) black cards and a set of \(6\) red cards. Each card has a number from \(1\) through \(6,\) s

by Brent@GMATPrepNow » Sat May 15, 2021 5:48 am
Vincen wrote: ↑
Thu May 13, 2021 6:09 am
 Bill has a set of \(6\) black cards and a set of \(6\) red cards. Each card has a number from \(1\) through \(6,\) such that each of the numbers \(1\) through \(6\) appears on \(1\) black card and \(1\) red card. Bill likes to play a game in which he shuffles all \(12\) cards, turns over \(4\) cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

(A) \(\dfrac8{33}\)

(B) \(\dfrac{62}{65}\)

(C) \(\dfrac{17}{33}\)

(D) \(\dfrac{103}{165}\)

(E) \(\dfrac{25}{33\})

Answer: C

We can solve this question using probability rules.

First, recognize that P(at least one pair) = 1 - P(no pairs)

P(no pairs) = P(select ANY 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33

Answer: C

Cheers,
Brent
Source: Manhattan GMAT
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1