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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

A fair sided die labeled 1 to 6 is tossed three times. What [#permalink]

Show Tags

09 Feb 2012, 08:02

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

75% (02:31) correct
25% (01:44) wrong based on 179 sessions

HideShow timer Statistics

A fair sided die labeled 1 to 6 is tossed three times. What is the probability the sum of the 3 throws is 16? A) 1/6 B) 7/216 C) 1/36 D) 9/216 E) 11/216

I have a problem with these type if questions, especially if the sum is sth like 12,etc. (the number of combinations are large)

What is the way to solve this problem?

could somebody solve this using the expected value concept please?

A fair sided die labeled 1 to 6 is tossed three times. What is the probability the sum of the 3 throws is 16? A) 1/6 B) 7/216 C) 1/36 D) 9/216 E) 11/216

I have a problem with these type if questions, especially if the sum is sth like 12,etc. (the number of combinations are large)

What is the way to solve this problem?

could somebody solve this using the expected value concept please?

I would say first remember that the probability of anything is the number of outcomes that give you what you want divided by the total number of possible outcomes.

Now, the total number of possible outcomes is, of course, 6^3.

Fortunately, there are only two combinations of three dice throws that will give you a sum of 16:

5, 5, 6 4, 6, 6

(I have a hard time remembering a real GMAT question that would give you 12, as the number of potential combinations is indeed inordinately large and difficult to map out.)

However, our denominator counts out the number of throwing sequences (permutations) and not just combinations, so we will need to do the same. Effectively there are three throwing sequences for either combination:

6, 5, 5 5, 6, 5 5, 5, 6

4, 6, 6 6, 4, 6 6, 6, 4

So 6 throwing sequences we want divided by 6^3 throwing sequences we could get gives us 1/(6^2) or 1/36.

As for the "expected value," that isn't, unfortunately, of much help on this particular problem since all of the denominators in the answer choices are powers of 6.
_________________

A fair sided die labeled 1 to 6 is tossed three times. What is the probability the sum of the 3 throws is 16? A) 1/6 B) 7/216 C) 1/36 D) 9/216 E) 11/216

I have a problem with these type if questions, especially if the sum is sth like 12,etc. (the number of combinations are large)

What is the way to solve this problem?

could somebody solve this using the expected value concept please?

Not sure what “expected value concept” is but used the following logic:

1. There are 2 ways to get 16 out of 3 throws: 6+6+4 and 6+5+5 2. There are 3 different ways to throw each of those combinations (6+6+4 or 6+4+6 or 4+6+6; the same is for 6-5-5), so we have 6 combinations to get 16 3. The probability of getting any number at a throw is 1/6, so the probability of throwing each combination is 1/6*1/6*1/6 = 1/216 4. 6*(1/216) = 1/36

A fair sided die labeled 1 to 6 is tossed three times. What is the probability the sum of the 3 throws is 16? A) 1/6 B) 7/216 C) 1/36 D) 9/216 E) 11/216

I have a problem with these type if questions, especially if the sum is sth like 12,etc. (the number of combinations are large)

What is the way to solve this problem?

could somebody solve this using the expected value concept please?

The sum of 16 is obtained if the dices show: (6, 5, 5) or (6, 6, 4).

Each can occur in 3!/2!=3 ways, for example we can get (6, 5, 5) in 3 ways: (6, 5, 5), (5, 6, 5), (5, 5, 6), permutation of three items (6, 6, 5) out of which 2 5's are identical.

So, # of favorable outcomes is 2*3=6; Total # of outcomes is 6*6*6=216;

P=favorable/total=6/216=1/36.

Answer: C.

As for your question: generally if on the real test you are faced with some hard calculations, something like 1.79*4.89, then it might indicate two cases: either there is a shortcut solution for this problem, which won't involve such tedious math or answer choices are handy to deal with it.

Now, 12 can be obtained by combination of the following 6 cases: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25 --> P=25/216. Since, it involves counting of 6 different cases and there is no real shortcut, then I doubt that you'll be asked about the probability of getting the sum of 12 (or the answers choices will be useful to find this value).

Having said that you should take into account the fact that the probability distribution in this question (and in many similar questions) is symmetrical and sometimes that could give a good shortcut. For example: The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum); The probability of getting the sum of 4 = the probability of getting the sum of 17; The probability of getting the sum of 5 = the probability of getting the sum of 16; ... The probability of getting the sum of 10 = the probability of getting the sum of 11;

So, for our case, the probability of getting the sum of 16 is the same as the probability of getting the sum of 5 and you can find the combinations for the sum 5 instead: (2, 2, 1) or (3, 1, 1).

To summarize: combinations/probability questions can be solve in many ways, most of them are fairly straightforward, some have tricky shortcut and some can be solved wit the help of answer choices, so no hard math will be required.

Check similar question which involves expected value approach to practice: mary-and-joe-126407.html

1=1 1+2 = 3 3+3 = 6 6+4 = 10 10+5 = 15 15+6 = 21 following two not in pattern. 21+4 = 25 25+2 = 27
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: A fair sided die labeled 1 to 6 is tossed three times. What [#permalink]

Show Tags

23 Aug 2014, 22:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: A fair sided die labeled 1 to 6 is tossed three times. What [#permalink]

Show Tags

23 Feb 2016, 13:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

A fair sided die labeled 1 to 6 is tossed three times. What is the probability the sum of the 3 throws is 16? A) 1/6 B) 7/216 C) 1/36 D) 9/216 E) 11/216

I have a problem with these type if questions, especially if the sum is sth like 12,etc. (the number of combinations are large)

What is the way to solve this problem?

could somebody solve this using the expected value concept please?

These two posts discuss the various possible sums:

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...