GRE Prep 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.
A box contains 12 balls, 7 of them are red and 5 of them are green. If
[#permalink]
04 Dec 2023, 00:58
04 Dec 2023, 00:58
Expert Reply
00:00
Question Stats:
50% (01:47) correct
50% (02:28) wrong based on 6 sessions
Hide Show timer Statistics
A box contains 12 balls, 7 of them are red and 5 of them are green. If 3 balls are to be selected at random from the box, what is the probability that 2 balls will be red and 1 ball will be green?
A. 7/44
B. 7/22
C. 51/100
D. 21/44
E. 7/9
A. 7/44
B. 7/22
C. 51/100
D. 21/44
E. 7/9
Part of the project: The Butler-GRE Daily New Quant and Verbal Questions to Practice (2023) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Manager
Joined: 11 Nov 2023
Posts: 203
Given Kudos: 76
WE:Business Development (Advertising and PR)
A box contains 12 balls, 7 of them are red and 5 of them are green. If
[#permalink]
04 Dec 2023, 19:46
04 Dec 2023, 19:46
1
1
Bookmarks
There are 3 different ways you could pick the balls: 1) Red, Red, Green vs. 2) Red, Green, Red vs. 3) Green, Red, Red.
Let's pick the first scenario. There are 7 out 12 Reds. Thus, the probability of picking a Red is 7/12. The probability of picking another Red is 6/11, since 6 Reds remain and there are now 11 total balls. The probability of picking a Green is 5/10.
\(\frac{7}{12} * \frac{6}{11} * \frac{5}{10}\)
\(= \frac{210}{1320}\)
\(= \frac{7}{44}\)
However, since there are 3 different ways, multiply by 3:
\(= \frac{7}{44}*3\)
\(= \frac{21}{44}\)
Let's pick the first scenario. There are 7 out 12 Reds. Thus, the probability of picking a Red is 7/12. The probability of picking another Red is 6/11, since 6 Reds remain and there are now 11 total balls. The probability of picking a Green is 5/10.
\(\frac{7}{12} * \frac{6}{11} * \frac{5}{10}\)
\(= \frac{210}{1320}\)
\(= \frac{7}{44}\)
However, since there are 3 different ways, multiply by 3:
\(= \frac{7}{44}*3\)
\(= \frac{21}{44}\)
gmatclubot
A box contains 12 balls, 7 of them are red and 5 of them are green. If [#permalink]
04 Dec 2023, 19:46
Moderators:
