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:

I am trying to solve this, but am getting it wrong.

What is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be a 6?

The way I am solving it:

Quote:

Since I need to find the probability that at least one of the rolls is a 6, there are three outcomes possible:

Outcome 1: Exactly one roll is a 6: This would mean one roll is a 6, while the remaining two are not. The probability would be: \(\frac{1}{6}*\frac{5}{6}*\frac{5}{6}\) \(= \frac{25}{216}\)

Outcome 2: Two rolls are 6s, one is not The probability would be: \(\frac{1}{6}*\frac{1}{6}*\frac{5}{6}\) \(= \frac{5}{216}\)

Outcome 3: All three rolls are 6s The probability would be: \(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}\) \(= \frac{1}{216}\)

Therefore, probability that at least one roll is a six would be Outcome 1 OR Outcome 2 OR Outcome 3

But the right answer is \(\frac{91}{216}\). The book has given this example to explain the 1-x probability trick, ie. calculating the probability of not getting even one 6 and subtracting it from 1.

I was trying to solve it the normal way, to understand it.

I seem to get 91/216 if I consider WHICH ROLL is a 6 in the first two outcomes, and add their probabilities. eg, for Outcome 1: First roll is a 6 AND the roll 2 and Roll 3 are not OR the Second roll is a 6 and the other two are not OR the third roll is a 6 and the other two are not

This would give me \(\frac{25}{216}+\frac{25}{216}+\frac{25}{216} = \frac{75}{216}\) for Outcome 1 and \(\frac{5}{216}+\frac{5}{216}+\frac{5}{216} = \frac{15}{216}\) for Outcome 2, Outcome 3 would be \(\frac{1}{216}\) for a total of \(\frac{91}{216}\)

But I don't see why WHICH ROLL is a 6 is important for this problem.

Re: Probability of at least one outcome [#permalink]

Show Tags

04 Apr 2013, 07:19

2

This post received KUDOS

What you say is correct, but not complete.

Case 1: only one 6. you here analyze one case (6,N,N) correctly, but you forget to analyze the other 2 cases : (N,6,N) and (N,N,6) so the probability of case one is \(3*\frac{25}{216}\)

The same happens for case 2: two 6s. Here again we have to multiply your result by 3 in order to get all the combinations: (N,6,6)(6,N,6)(6,6,N) \(3*\frac{5}{216}\)

The probability of the third case is correct. Now we sum our results and obtain \(\frac{91}{216}\)
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Probability of at least one outcome [#permalink]

Show Tags

04 Apr 2013, 08:26

Zarrolou wrote:

What you say is correct, but not complete.

Case 1: only one 6. you here analyze one case (6,N,N) correctly, but you forget to analyze the other 2 cases : (N,6,N) and (N,N,6) so the probability of case one is \(3*\frac{25}{216}\)

The same happens for case 2: two 6s. Here again we have to multiply your result by 3 in order to get all the combinations: (N,6,6)(6,N,6)(6,6,N) \(3*\frac{5}{216}\)

The probability of the third case is correct. Now we sum our results and obtain \(\frac{91}{216}\)

Thanks Zarrolou.

I'm having trouble understanding the concept of probability. Im trying to frame this question as a Combinatorics question. Please help me out.

Lets just consider Case 1, Only one 6 in three rolls.

So, for three rolls, all the possible outcomes are:

6 outcomes of first roll AND 6 of second AND 6 of third = 216

So, the right way to frame the question now would be : What are the different ways of getting exactly one 6 in three rolls of a dice?

Using the Slots method, that would be:

1 AND 5 AND 5 OR 5 AND 1 AND 5 OR 5 AND 5 AND 1.

That would be 75.

So, probability of getting exactly one 6 would now be, \(\frac{Ways of getting desired outcome}{All possible outcomes}\) = \(\frac{75}{216}\)

I am sorry for taking this roundabout rote, but am just trying to get it clear for myself.

Re: Probability of at least one outcome [#permalink]

Show Tags

04 Apr 2013, 09:27

1

This post received KUDOS

rooker wrote:

Thanks Zarrolou.

I'm having trouble understanding the concept of probability. Im trying to frame this question as a Combinatorics question. Please help me out.

Lets just consider Case 1, Only one 6 in three rolls.

So, for three rolls, all the possible outcomes are:

6 outcomes of first roll AND 6 of second AND 6 of third = 216

So, the right way to frame the question now would be : What are the different ways of getting exactly one 6 in three rolls of a dice?

Using the Slots method, that would be:

1 AND 5 AND 5 OR 5 AND 1 AND 5 OR 5 AND 5 AND 1.

That would be 75.

So, probability of getting exactly one 6 would now be, \(\frac{Ways of getting desired outcome}{All possible outcomes}\) = \(\frac{75}{216}\)

I am sorry for taking this roundabout rote, but am just trying to get it clear for myself.

Absolutely correct! The probability of getting one 6 and three N6s is: \(\frac{1}{6}*\frac{5}{6}*\frac{5}{6}\) The number of possible outcome is which this can happen is 3. So \(3*\frac{25}{216}\) is the magic number
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Probability of at least one outcome [#permalink]

Show Tags

12 Apr 2013, 01:28

1

This post received KUDOS

Hi there,

This is my way of approach: The probability of not getting a 6 in 3 throws is (5/6)X(5/6)x(5/6). Therefore the probability of getting at least one 6 is 1- [P(no 6 in 3 throws) = (5/6)X(5/6)x(5/6)] = 91/216

gmatclubot

Re: Probability of at least one outcome
[#permalink]
12 Apr 2013, 01:28

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

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...