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.
05 Oct 2004, 22:21
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
True story--and good GMAT problem-
friend in high school years ago took a 100 question multiple choice test.
Each question has answer options A,B,C,D, and E. So 5 choices.
He scored a 61% on the test by filling in "ACDC" for answers
1, 2,3, and 4, respectively, and he repeated this till the end of the test.
so 25 sets of ACDC. incredible! (yes, ACDC after the Australian band.)
What is the probability of this?
pretty simple, right?
friend in high school years ago took a 100 question multiple choice test.
Each question has answer options A,B,C,D, and E. So 5 choices.
He scored a 61% on the test by filling in "ACDC" for answers
1, 2,3, and 4, respectively, and he repeated this till the end of the test.
so 25 sets of ACDC. incredible! (yes, ACDC after the Australian band.)
What is the probability of this?
pretty simple, right?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
06 Oct 2004, 12:32
Kudos
Bookmarks
What the heck...I'll take a crack at it.
First off, I'm assuming that the answer choice is randomly generating (i.e. there is always a 1/5 chance that any random answer picked is correct).
Now that the assumptions are out of the way:
The chance of getting the 1st 61 questions right and the last 39 questions wrong is (1/5)^61 * (4/5)^39 = 3.831 * 10^-47.
Next: figure out the number of combinations that you can get 61 questions right and 39 questions wrong.
100!/61!39! = 9.014*10^27
We want to find p(combo1) OR p(combo2) OR p(combo3)...OR P(combo 9*10^27)
Since all combos have the same liklihood of occuring, the probability of scoring 61% on a 100 question test using random guesses (guessing ACDC without looking at the question should have the same probability of guessing all As or Bs) would be 3.831*10^-47 * 9.014 *10^27 = 3.453 *10^-19.
In other words there would be a 0.00000000000000003453% of getting 61 right on a 100 question test where the right answer choices are randomly and all answers are bubbled in randomly.
Anyone care to share their ideas?
First off, I'm assuming that the answer choice is randomly generating (i.e. there is always a 1/5 chance that any random answer picked is correct).
Now that the assumptions are out of the way:
The chance of getting the 1st 61 questions right and the last 39 questions wrong is (1/5)^61 * (4/5)^39 = 3.831 * 10^-47.
Next: figure out the number of combinations that you can get 61 questions right and 39 questions wrong.
100!/61!39! = 9.014*10^27
We want to find p(combo1) OR p(combo2) OR p(combo3)...OR P(combo 9*10^27)
Since all combos have the same liklihood of occuring, the probability of scoring 61% on a 100 question test using random guesses (guessing ACDC without looking at the question should have the same probability of guessing all As or Bs) would be 3.831*10^-47 * 9.014 *10^27 = 3.453 *10^-19.
In other words there would be a 0.00000000000000003453% of getting 61 right on a 100 question test where the right answer choices are randomly and all answers are bubbled in randomly.
Anyone care to share their ideas?
