GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 22 Oct 2018, 03:02

Close

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

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.

Close

Request Expert Reply

Confirm Cancel

What is the probability of rolling three fair dice and having at least

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50039
What is the probability of rolling three fair dice and having at least  [#permalink]

Show Tags

New post 31 Jan 2017, 10:02
1
3
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

76% (01:20) correct 24% (01:37) wrong based on 238 sessions

HideShow timer Statistics

Manager
Manager
avatar
S
Joined: 25 Nov 2016
Posts: 54
Location: Switzerland
GPA: 3
Premium Member
Re: What is the probability of rolling three fair dice and having at least  [#permalink]

Show Tags

New post 31 Jan 2017, 13:57
1
2
We have to (easier this way) calculate first the prob to have none even number (NEN).
We have a probability of 1/2 to have an even number (2,4,6)

=> Prob(NEN) = 1/2 * 1/2 * 1/2 = 1/8

=> Prob(at least one even number) = 1- Prob(NEN) = 1-1/8 = 7/8
CEO
CEO
User avatar
D
Joined: 12 Sep 2015
Posts: 3024
Location: Canada
Re: What is the probability of rolling three fair dice and having at least  [#permalink]

Show Tags

New post 03 Feb 2017, 08:33
Top Contributor
2
Bunuel wrote:
What is the probability of rolling three fair dice and having at least one of the three dice show an even number?

A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216


We want P(select at least 1 even number)
When it comes to probability questions involving "at least," it's usually best to try using the complement.

That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 even number) = 1 - P(not getting at least 1 even number)
What does it mean to not get at least 1 even number?
It means getting zero even numbers.
So, we can write: P(getting at least 1 even number) = 1 - P(getting zero even numbers)

P(getting zero even numbers) = P(1st coin is odd AND 2nd coin is odd AND 3rd coin is odd )
= P(1st coin is odd) x P(2nd coin is odd) x P(3rd coin is odd )
= 1/2 x 1/2 x 1/2
= 1/8

So, P(getting at least 1 even number) = 1 - 1/8
= 7/8

Answer: B
_________________

Brent Hanneson – GMATPrepNow.com
Image
Sign up for our free Question of the Day emails

Target Test Prep Representative
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3907
Location: United States (CA)
Re: What is the probability of rolling three fair dice and having at least  [#permalink]

Show Tags

New post 03 Feb 2017, 12:47
Bunuel wrote:
What is the probability of rolling three fair dice and having at least one of the three dice show an even number?

A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216


We can look at this problem in terms of only 2 possible events. Either at least one die shows an even number or no dice show an even number. This means:
P(at least one die showing an even number) + P(no dice showing an even number) = 1

P(at least one die showing an even number) = 1 - P(no dice showing an even number)

Thus, if we can determine the probability of no dice showing an even number, we’ll quickly be able to calculate the probability that at least one die shows an even number.

There are 3 even and 3 odd numbers on each die. Thus, the probability of not showing an even number on all three dice is 1/2 x 1/2 x 1/2 = 1/8.

Thus, the probability of showing an even number on at least one die is 1 = 1/8 = 7/8.

Answer: B
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Manager
Manager
avatar
B
Joined: 18 Jun 2017
Posts: 60
Re: What is the probability of rolling three fair dice and having at least  [#permalink]

Show Tags

New post 03 Aug 2017, 02:59
P (at least one even) = 1 - P(all odd) = 1[Universe]−1/2∗1/2∗1/2=1−1/8=7/8. [Probability of all odd'd on rolling 3 dice is 1/2∗1/2∗1/2=1/8]
Option B.
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 8515
Premium Member
Re: What is the probability of rolling three fair dice and having at least  [#permalink]

Show Tags

New post 17 Sep 2018, 19:35
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Bot
Re: What is the probability of rolling three fair dice and having at least &nbs [#permalink] 17 Sep 2018, 19:35
Display posts from previous: Sort by

What is the probability of rolling three fair dice and having at least

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.