BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Probability of receiving a five or a one on either die,

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members
Thanked: 11 times
by gmattesttaker2 » Sun May 04, 2014 8:32 pm
Hello,

Can you please assist with this:

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


OA: [spoiler]5/9[/spoiler]


Thanks a lot,
Sri
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sun May 04, 2014 10:07 pm
gmattesttaker2 wrote: In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
So, the player wins if he/she rolls AT LEAST one 5 or 1.
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= [spoiler]5/9[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members
Thanked: 11 times

by gmattesttaker2 » Mon May 05, 2014 6:00 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
So, the player wins if he/she rolls AT LEAST one 5 or 1.
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= [spoiler]5/9[/spoiler]

Cheers,
Brent

Hello Brent,

Thank you very much for your excellent and detailed explanation.

Best Regards,
Sri
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Mon May 05, 2014 10:43 pm
Hi Sri,

Brent's solution is great (and it's exactly how I would approach the question). There is another way to approach it though. Since you're rolling 2 dice, there aren't that many possible outcomes (just 36 in total), so you COULD use brute force and 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

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members
Thanked: 11 times

by gmattesttaker2 » Tue May 06, 2014 7:04 pm
[email protected] wrote:Hi Sri,

Brent's solution is great (and it's exactly how I would approach the question). There is another way to approach it though. Since you're rolling 2 dice, there aren't that many possible outcomes (just 36 in total), so you COULD use brute force and 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

GMAT assassins aren't born, they're made,
Rich
Hello Rich,

Thanks a lot for the alternate approach.

Best Regards,
Sri
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members
Thanked: 11 times

by gmattesttaker2 » Wed May 07, 2014 8:03 am
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
So, the player wins if he/she rolls AT LEAST one 5 or 1.
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= [spoiler]5/9[/spoiler]

Cheers,
Brent

Hello Brent,

I was just wondering if we can apply the same technique of

P(Event A happening) = 1 - P(Event A not happening)

for the following as well or should this technique be strictly applied for only the ones that have "at least".


Mathematics, physics, and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many mathematics books as physics books and the number of physics books is 4 greater than that of the chemistry books. Among all the books, 12 books are softcover and the remaining are hard-cover. If there are a total of 7 hard-cover books among the mathematics and physics books, what is the probability that a book selected at random is either a hard-cover book or a chemistry book?

OA: [spoiler]9/20[/spoiler]

m + p + c = 80/100(25)
=> m + p + c = 20

m = 2p and p = 4 + c
=> 2p + p + c = 20
=> 3p + c = 20
=> 3(4 + c) + c = 20
=> 12 + 3c + c = 20
=> 12 + 4c = 20
=> 4c = 8
=> c = 2
=> m + p = 18

Total Hard covers = 8
mathematics Hard cover + physics Hard cover = 7 => chemistry Hard cover = 1
Also, this means Total Soft Covers = 12


The probability that a book selected at random is either a hard-cover book or a chemistry book
= 1 - P (Not selecting a book that is either hard cover or chemistry)

However, I was not very sure how to solve from this point onwards.

Thanks a lot for your help.

Best Regards,
Sri
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Sat Mar 28, 2015 5:55 am

by e30sport » Fri Jun 12, 2015 8:07 am
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
So, the player wins if he/she rolls AT LEAST one 5 or 1.
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= [spoiler]5/9[/spoiler]

Cheers,
Brent
Brent,

Please explain to me why it's not just P(A and B) = P(A) x P(B). You can have two outcomes on the first and two on the second which would be 2/6 * 2/6.

I clearly don't understand the principles.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 154
Joined: Wed May 21, 2014 4:29 amFollowed by: 1 member
Thanked: 8 times

by talaangoshtari » Fri Jun 12, 2015 11:59 am
p(1,1) = 1/6 × 1/6 = 1/36
p(1,2) = 1/6 × 1/6 = 1/36
p(1,3) = 1/6 × 1/6 = 1/36
p(1,4) = 1/6 × 1/6 = 1/36
p(1,5) = 1/6 × 1/6 = 1/36
p(1,6) = 1/6 × 1/6 = 1/36

p(2,1) = 1/6 × 1/6 = 1/36
p(2,5) = 1/6 × 1/6 = 1/36
p(3,1) = 1/6 × 1/6 = 1/36
p(3,5) = 1/6 × 1/6 = 1/36
p(4,1) = 1/6 × 1/6 = 1/36
p(4,5) = 1/6 × 1/6 = 1/36

p(5,1) = 1/6 × 1/6 = 1/36
p(5,2) = 1/6 × 1/6 = 1/36
p(5,3) = 1/6 × 1/6 = 1/36
p(5,4) = 1/6 × 1/6 = 1/36
p(5,5) = 1/6 × 1/6 = 1/36
p(5,6) = 1/6 × 1/6 = 1/36
p(6,1) = 1/6 × 1/6 = 1/36
p(6,5) = 1/6 × 1/6 = 1/36

Therefore, the total sum is 20/36. So, there is more than two outcomes for the first and second, 2/6 does not account for all the possibilities.
Last edited by talaangoshtari on Fri Jun 12, 2015 1:00 pm, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Fri Jun 12, 2015 12:50 pm
e30sport wrote:
Brent,

Please explain to me why it's not just P(A and B) = P(A) x P(B). You can have two outcomes on the first and two on the second which would be 2/6 * 2/6.

I clearly don't understand the principles.
Hi e30sport,

Applying P(A and B) means finding the probability that we get 5 or 1 on the first die AND 5 or 1 on the second die.
To win the game we don't need to get 5 or 1 on BOTH dice.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Thu Aug 13, 2015 7:56 am

by Anantjit » Wed Jul 27, 2016 5:55 am
It can be like below :-
Probability of 5 or 1 on either dice= P(5) or P(1) on 1st dice OR P(5) or P(1) on 2nd Dice
i.e P[A]+P-P[A and B]
P[A]=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
P=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
P[A and B]= 1/3+1/3-[1/3*1/3]=2/3-1/9=5/9
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members
Location: East Bay all the way
Thanked: 625 times
GMAT Score:780

by Matt@VeritasPrep » Thu Aug 04, 2016 9:16 pm
Anantjit wrote:It can be like below :-
Probability of 5 or 1 on either dice= P(5) or P(1) on 1st dice OR P(5) or P(1) on 2nd Dice
i.e P[A]+P-P[A and B]
P[A]=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
P=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
P[A and B]= 1/3+1/3-[1/3*1/3]=2/3-1/9=5/9


This is the right idea, but you want to minimize the amount of clutter on your page during the test, or it's easy to confuse yourself with too much superfluous jargon. (It also probably wouldn't help someone who doesn't understand the idea in the first place, as it's a bit of a wall of text: keeping it simple and clean is good for the teacher and the student :D)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1