Veritas Prep
What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (\(52\) cards, with \(13\) sets of four matching suits)?
A. \(\dfrac{12}{2,652}\)
B. \(\dfrac{16}{2,652}\)
C. \(\dfrac{1}{17}\)
D. \(\dfrac{1}{13}\)
E. \(\dfrac{1}{2}\)
OA C
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards
This topic has expert replies
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
P(select pair) = P(1st card is ANY card AND 2nd card matches 1st card)AAPL wrote: ↑Tue May 09, 2023 11:46 amVeritas Prep
What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (\(52\) cards, with \(13\) sets of four matching suits)?
A. \(\dfrac{12}{2,652}\)
B. \(\dfrac{16}{2,652}\)
C. \(\dfrac{1}{17}\)
D. \(\dfrac{1}{13}\)
E. \(\dfrac{1}{2}\)
OA C
= P(1st card is ANY card) x P(2nd card matches 1st card)
= 1 x 3/51
= 3/51
= 1/17
= C
Aside: P(2nd card matches 1st card) = 3/51, because once 1 card is selected, there are 51 cards remaining in the deck. Among those 51 remaining cards, there are 3 that match the 1st card selected.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
