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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
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.
10 Apr 2018, 22:43
Kudos
Bookmarks
Bunuel
I understand. But the question doesn't mention a time frame.
For example, considering the probability to be 3/4 is for any 4 year time frame. If we consider a 5 year time frame, the probabilities change.
The probabilities you asked depend on other factors. One can simply not tell. But with no other information given, the probability that I could meet a dinosaur would be 1/2. I may or may not encounter a dinosaur.
Please help.
Thanks.
10 Apr 2018, 23:35
Kudos
Bookmarks
Bakshi121092
That's not correct.
There is one leap year in four. So, the probability of random person being born in leap year is 1/4. As for the dinosaur problem, no other information is needed apart from common knowledge, the probability is not 1/2: meeting and not meeting are not equally likely, and simply the fact that there are only two options does not mean that the provability is 1/2.
I suggest you to follow the links given in my previous post and maybe read some simple book on probability (introduction of some kind).
24 Sep 2024, 00:24
Kudos
Bookmarks
desaichinmay22
I don't think, this is the right approach.
Probability of a person to be born in a leap year = 1/4 ... <1/2 (alright)
Probability of at least one person out of two to be born in a leap year = 1/4 x 3/4 +3/4 x 1/4 + 1/4 x 1/4 = 7/16 <0.5
Probability of at least one person out of 3 to be born in a leap year = 1 - negative of no one is born in leap year = 1- (3/4) x (3/4) x (3/4) = 37/64 >0.5
hence answer is C
It is better to use non event method in such questions. Reduces lot of efforts and saves time
