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) \(\dfrac{8}{33}\)
(B) \(\dfrac{62}{65}\)
(C) \(\dfrac{17}{33}\)
(D) \(\dfrac{103}{165}\)
(E) \(\dfrac{25}{33}\)
Answer: C
Source: Manhattan GMAT
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....
This topic has expert replies
-
- Legendary Member
- Posts: 1622
- Joined: Thu Mar 01, 2018 7:22 am
- Followed by:2 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
We can solve this question using probability rules.Gmat_mission wrote: ↑Sun Sep 12, 2021 9:34 amBill 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) \(\dfrac{8}{33}\)
(B) \(\dfrac{62}{65}\)
(C) \(\dfrac{17}{33}\)
(D) \(\dfrac{103}{165}\)
(E) \(\dfrac{25}{33}\)
Answer: C
Source: Manhattan GMAT
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