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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
24 May 2012, 00:52

4

This post received KUDOS

Expert's post

cyberjadugar wrote:

Gambling with 4 dice, what is the probability of getting an even sum?

A. 3/4 B. 1/2 C. 2/3 D. 1/4 E. 1/3

Looking for a one-liner answer...

Even sum can be obtained in following cases: EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4; OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4; EEOO - \frac{4!}{2!2!}=6 cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}.

Answer: B.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 1/2.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
24 May 2012, 01:26

Hi Bunuel,

The shortcut is perfect! Are you aware of any thread where we such approaches are discussed?

Similar solution can be applied to problem like no. of cases A occurs before B in a sequence, no. of cases where A & B are together In both cases half of the total no. of cases would be the answer. Since, either A comes after B or before B, either A & B are together or separated.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
24 May 2012, 01:30

Expert's post

cyberjadugar wrote:

Hi Bunuel,

The shortcut is perfect! Are you aware of any thread where we such approaches are discussed?

Similar solution can be applied to problem like no. of cases A occurs before B in a sequence, no. of cases where A & B are together In both cases half of the total no. of cases would be the answer. Since, either A comes after B or before B, either A & B are together or separated.

Hey everyone, today’s post focuses on the interview process. As I get ready for interviews at Kellogg and Tuck (and TheEngineerMBA ramps up for his HBS... ...

I got invited to interview at Sloan! The date is October 31st. So, with my Kellogg interview scheduled for this Wednesday morning, and my MIT Sloan interview scheduled...