|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
15 Jan 2023, 03:20
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
51% (02:00) correct
49%
(02:34)
wrong
based on 93
sessions
History
Date
Time
Result
Not Attempted Yet
A black sack contains three green balls, five yellow balls, and four white ones. Three balls are taken at random without replacement. What is the probability of having all the the three balls of different colors?
A. 3/11
B. 3/22
C. 1/22
D. 1/33
E. 1/121
A. 3/11
B. 3/22
C. 1/22
D. 1/33
E. 1/121
Updated on: 18 Jan 2023, 10:17
Originally posted by MimRahman452 on 15 Jan 2023, 11:42.
Last edited by MimRahman452 on 18 Jan 2023, 10:17, edited 1 time in total.
Last edited by MimRahman452 on 18 Jan 2023, 10:17, edited 1 time in total.
Kudos
Bookmarks
(3C1 * 12C1 ) * (5C1 * 11C1 ) * (4C1 * 10C1 ) * 3! = 3/11
ANS. = A
ANS. = A
15 Jan 2023, 12:53
Kudos
Bookmarks
Bunuel
We need three balls of different color, let's say we draw the ball in the order of green - yellow and white
Probability of drawing a green ball = 3/12
Probability of drawing a yellow ball after a green ball has been drawn = 5/11
Probability of drawing a white ball after a green ball and a yellow ball have been drawn = 4/10
Net probability = 3/12 * 5/11 * 4/10
Now the balls can be drawn in any order. Number of ways the balls can be drawn = 3!
Required probability = 3/12 * 5/11 * 4/10 * 3!
= 3/11
Option A
