Events & Promotions
|
|
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.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
jcaguirre91
Joined: 20 Oct 2020
Last visit: 26 Feb 2025
Posts: 19
Given Kudos: 1,203
Location: Mexico
Schools: HBS '25 (D) Booth '25 (I) Sloan '25 (WL)
GMAT 1: 690 Q48 V35
GPA: 3.8
15 Oct 2021, 19:16
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:11) correct
40%
(02:28)
wrong
based on 450
sessions
History
Date
Time
Result
Not Attempted Yet
In a class, every student is either American or European. Every student is wearing a blazer or a sweater. If a student is selected randomly, is the probability of the student being a European wearing a sweater greater than that of his/her being an American wearing a blazer?
(1) The probability that a randomly selected student is either American or wearing a blazer is 5/6.
(2) The probability that a randomly selected student is either a European or wearing a sweater is 3/5.
(1) The probability that a randomly selected student is either American or wearing a blazer is 5/6.
(2) The probability that a randomly selected student is either a European or wearing a sweater is 3/5.
16 Oct 2021, 22:00
Kudos
Bookmarks
In a class, every student is either American or European. Every student is wearing a blazer or a sweater. If a student is selected randomly, is the probability of the student being a European wearing a sweater greater than that of his/her being an American wearing a blazer?
You can make a 2 by 2 matrix or take variables.
There are four categories of people: \(T=E_b + E_s +A_b+ A_s\)
We are comparing the probability of two categories
Thus we are looking for: Whether \(\frac{E_s}{T}>\frac{A_b}{T}\)
(1) The probability that a randomly selected student is either American or wearing a blazer is 5/6.
\(\frac{E_b+A_s+A_b}{T}=\frac{5}{6}\)
\(1-\frac{E_b+A_s+A_b}{T}=1-\frac{5}{6}\)
\(\frac{T-(E_b+A_s+A_b)}{T}=\frac{6-5}{6}\)
\(\frac{E_s}{T}=\frac{1}{6}\)
Nothing about \(A_b\)
(2) The probability that a randomly selected student is either a European or wearing a sweater is 3/5.
\(\frac{E_b+E_s+A_s}{T}=\frac{3}{5}\)
\(1-\frac{E_b+E_s+A_s}{T}=1-\frac{3}{5}\)
\(\frac{T-(E_b+E_s+A_s)}{T}=\frac{2}{5}\)
\(\frac{A_b}{T}=\frac{2}{5}\)
Nothing about \(E_s\)
Combined
We can say \(\frac{2}{5}>\frac{1}{6}\)
Thus, \(P(A_b)>P(E_s)\)
Sufficient
C
You can make a 2 by 2 matrix or take variables.
There are four categories of people: \(T=E_b + E_s +A_b+ A_s\)
We are comparing the probability of two categories
Thus we are looking for: Whether \(\frac{E_s}{T}>\frac{A_b}{T}\)
(1) The probability that a randomly selected student is either American or wearing a blazer is 5/6.
\(\frac{E_b+A_s+A_b}{T}=\frac{5}{6}\)
\(1-\frac{E_b+A_s+A_b}{T}=1-\frac{5}{6}\)
\(\frac{T-(E_b+A_s+A_b)}{T}=\frac{6-5}{6}\)
\(\frac{E_s}{T}=\frac{1}{6}\)
Nothing about \(A_b\)
(2) The probability that a randomly selected student is either a European or wearing a sweater is 3/5.
\(\frac{E_b+E_s+A_s}{T}=\frac{3}{5}\)
\(1-\frac{E_b+E_s+A_s}{T}=1-\frac{3}{5}\)
\(\frac{T-(E_b+E_s+A_s)}{T}=\frac{2}{5}\)
\(\frac{A_b}{T}=\frac{2}{5}\)
Nothing about \(E_s\)
Combined
We can say \(\frac{2}{5}>\frac{1}{6}\)
Thus, \(P(A_b)>P(E_s)\)
Sufficient
C
31 Jul 2023, 14:14
Kudos
Bookmarks
jcaguirre91
We can also use a 2 * 2 matrix to plot the information.
Attachment:
1.jpg [ 13.95 KiB | Viewed 6766 times ]
AMR ⇒ Students who are Americans
EUR ⇒ Students who are Europeans
BLZ ⇒ Students who wear blazers
SWT ⇒ Students who wear sweaters
We need to find if \(x \geq y\)
Statement 1
(1) The probability that a randomly selected student is either American or wearing a blazer is 5/6.
Attachment:
2.jpg [ 16.3 KiB | Viewed 6639 times ]
We have been presented with the combined probability of a student who is a part of the blocks indicated by a red cross.
The sum of the probability of all four boxes equals 1
Probability of a European wearing a sweater = \(1-\frac{5}{6} = \frac{1}{6}\)
However, we do not have any information on the probability of an American wearing a blazer. Hence, this statement alone is not sufficient.
Statement 2
Attachment:
3.jpg [ 17.66 KiB | Viewed 6479 times ]
(2) The probability that a randomly selected student is either a European or wearing a sweater is 3/5
We have been presented with the combined probability of a student who is a part of the blocks indicated by a green cross.
The sum of the probability of all four boxes equals 1
Probability of an American wearing a blazer = \(1-\frac{3}{5} = \frac{2}{5}\)
However, we do not have any information on the probability of a European wearing a sweater. Hence, this statement alone is not sufficient.
Combined
Attachment:
4.jpg [ 14.83 KiB | Viewed 6464 times ]
As we have both probabilities, we can compare the values and obtain a definite answer to the question. The statements combined are sufficient.
Option C
