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 350,000 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:

In a game, one player throws two fair, six-sided die at the [#permalink]
01 May 2013, 10:03

2

This post received KUDOS

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

35% (02:31) correct
65% (01:29) wrong based on 161 sessions

In a game, one player throws two fair, six-sided die at the same time. If the player receives at least a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
01 May 2013, 10:40

Option C.

the number of cases in which he can lose the game are when both the faces have neither of 5 or 1 or both. so the possible combinations are (2,2),(2,3),(2,4) (2,6) and 12 more with 3,4,6.

probability of loss = # loss cases/# total no of cases = 16/36 or 4/9

hence probability of win = 1-p(loss). = 1-(4/9) = 5/9

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
01 May 2013, 10:41

2

This post received KUDOS

2

This post was BOOKMARKED

We have 2 good (read five and one) possibilities (\(G\)) on 6 faces G=2/6 and 4 bad possibilities (\(B\)) on 6 faces B=4/6 The winning combinations are the ones with at least a \(G\) in it so: \(G,B\) \(B,G\) \(G,G\)

\(G,B\) and \(B,G\) have the same probability \(\frac{2}{6}*\frac{4}{6}=\frac{2}{9}\) each \(G,G\) has a probability of \(\frac{2}{6}*\frac{2}{6}=\frac{1}{9}\) Sum them up \(\frac{2}{9}+\frac{2}{9}+\frac{1}{9}=\frac{5}{9}\) _________________

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

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
02 May 2013, 03:24

6

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

karishmatandon wrote:

In a game, one player throws two fair, six-sided die at the same time. If the player receives at least a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?

A. 1/3 B. 4/9 C. 5/9 D. 2/3 E. 3/4

Probably the easiest approach would be to find the probability of the opposite event and subtract it from 1:

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
05 May 2013, 04:59

Zarrolou wrote:

We have 2 good (read five and one) possibilities (\(G\)) on 6 faces G=2/6 and 4 bad possibilities (\(B\)) on 6 faces B=4/6 The winning combinations are the ones with at least a \(G\) in it so: \(G,B\) \(B,G\) \(G,G\)

\(G,B\) and \(B,G\) have the same probability \(\frac{2}{6}*\frac{4}{6}=\frac{2}{9}\) each \(G,G\) has a probability of \(\frac{2}{6}*\frac{2}{6}=\frac{1}{9}\) Sum them up \(\frac{2}{9}+\frac{2}{9}+\frac{1}{9}=\frac{5}{9}\)

WHy in both winning combination we are calculating for GG only once . May be on first die 5 and second die one or on first die one and second die 5...These can combinations can also occur na? ..WHy we are not considering this scenario? _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
05 May 2013, 05:43

1

This post received KUDOS

skamal7 wrote:

WHy in both winning combination we are calculating for GG only once . May be on first die 5 and second die one or on first die one and second die 5...These can combinations can also occur na? ..WHy we are not considering this scenario?

Hi skamal7,

Consider the following example that will explain better than any theoretical information. You say that G,G should be counted twice, so the possible combinations are:

G,G=1/9 G,G=1/9 B,G=2/9 G,B=2/9 B,B=4/9 [ also if your method is correct B,B should be counted twice =4/9 ]

don't you see anything odd? The sum of the probability of each case is greater than 1! \(\frac{1+1+2+2+4}{9}=\frac{10}{9}\) [ if you count B,B twice it becomes \(\frac{14}{9}\) ]

Why does this happen?Let's look at the theory now The formula to solve this problem is \((nCk)p^k*q^{(n-k)}\) where p=1/3 and q=2/3 and N are the dies and K are the good outcomes:

Case two good \((2C2)(\frac{1}{3})^2(\frac{2}{3})^0=\frac{1}{9}\) Case one good one bad \((2C1)(\frac{1}{3})^1(\frac{2}{3})^1=\frac{4}{9}\) Case two bad \((2C0)(\frac{1}{3})^0(\frac{2}{3})^2=\frac{4}{9}\)

Tot sum = \(\frac{1+4+4}{9}=1\)

Hope it's clear now, let me know _________________

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

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
05 May 2013, 08:09

1

This post received KUDOS

karishmatandon wrote:

In a game, one player throws two fair, six-sided die at the same time. If the player receives at least a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?

A. 1/3 B. 4/9 C. 5/9 D. 2/3 E. 3/4

Instead of trying to count the overlapping events and thereby complicating the probability calculation, we can simply calculate the probability of 'not winning' and subtract it from 1 to get the probability of 'winning'

Therefore required probability \(P = 1 - (\frac{4}{6})*(\frac{4}{6})\)

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
30 Jul 2013, 10:51

I also got confused with this at least, i interpreted it as 5-x 6-x 1-x 5-1 6-1 vice-versa cases. _________________

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: In a game, one player throws two fair, six-sided die at the [#permalink]
03 Aug 2013, 03:10

dyaffe55 wrote:

I think the question should be re-worded. 'At least a five' sounds like >= 5. Therefore, my result was 1-(1/2*1/2) = 3/4

I made the same mistake as well... But for the condition above wouldnt the answer be 5/6?

Bunuel/Zarroulou could you confirm? If a win was 1,5,6 instead of 1 and 5? _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
03 Aug 2013, 03:52

I meant the condition for a win was at least a five (5 or 6) or 1 on either die... _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
23 Mar 2014, 16:14

1

This post received KUDOS

Question is not clear when it refers to 'at least' does it refer to at least a 5, that meaning 5 or 6? Or does it refer to have at least one of two: either a 5 or a 1?

Re: In a game, one player throws two fair, six-sided die at the [#permalink]
20 Jun 2015, 11:24

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: In a game, one player throws two fair, six-sided die at the [#permalink]
20 Jun 2015, 17:10

Expert's post

Hi All,

This question can be solved with "brute force." Since you're rolling 2 dice, there aren't that many possible outcomes (just 36 in total), so you COULD just write them all down:

We're looking for the number of outcomes that include AT LEAST a 1 or a 5.

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,5

3,1 3,5

4,1 4,5

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,5

Total possibilities = 20

Probability of rolling at least a 1 or a 5 on two dice: 20/36 = 5/9

Harvard asks you to write a post interview reflection (PIR) within 24 hours of your interview. Many have said that there is little you can do in this...