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 exactly 22 balls, of which 11 are white and 11 are blac
[#permalink]
04 Dec 2023, 00:57
04 Dec 2023, 00:57
Expert Reply
00:00
Question Stats:
100% (01:00) correct
0% (00:00) wrong based on 9 sessions
Hide Show timer Statistics
A box contains exactly 22 balls, of which 11 are white and 11 are black. If two balls are to be picked from this box at random and without replacement, what is the probability that both balls will be black?
A. 1/6
B. 5/21
C. 5/11
D. 11/40
E. 17/40
A. 1/6
B. 5/21
C. 5/11
D. 11/40
E. 17/40
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: 214
Given Kudos: 76
WE:Business Development (Advertising and PR)
Re: A box contains exactly 22 balls, of which 11 are white and 11 are blac
[#permalink]
04 Dec 2023, 20:46
04 Dec 2023, 20:46
1
There are 11 Black balls out 22 total balls. So, the probability of picking a Black the first time is 11/22. Now we have 10 Black balls out of 21 total balls. The probability of picking a Black is now 10/21.
\(\frac{11}{22} * \frac{10}{21}\)
\(= \frac{1}{2}*\frac{10}{21}\)
\(= \frac{5}{21}\)
\(\frac{11}{22} * \frac{10}{21}\)
\(= \frac{1}{2}*\frac{10}{21}\)
\(= \frac{5}{21}\)
A box contains exactly 22 balls, of which 11 are white and 11 are blac
[#permalink]
30 Jan 2024, 19:42
30 Jan 2024, 19:42
1
The probability of picking the black ball the first time is
\(\dfrac{11}{22}\)
The probability of picking the black ball the second times without replacement is
\(\dfrac{10}{21}\)
The probability that both balls will be black is
\(\dfrac{11}{22} \times \dfrac{10}{21} = \dfrac{1}{2} \times \dfrac{10}{21} = \dfrac{5}{21}\)
The answer is Choice B.
\(\dfrac{11}{22}\)
The probability of picking the black ball the second times without replacement is
\(\dfrac{10}{21}\)
The probability that both balls will be black is
\(\dfrac{11}{22} \times \dfrac{10}{21} = \dfrac{1}{2} \times \dfrac{10}{21} = \dfrac{5}{21}\)
The answer is Choice B.
