It is currently 16 Dec 2017, 01:28

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Crowan throws 3 dice and records the product of the numbers

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

Hide Tags

1 KUDOS received
Manager
Manager
avatar
Joined: 16 Oct 2009
Posts: 206

Kudos [?]: 44 [1], given: 8

Schools: HEC Paris, , Tepper
Crowan throws 3 dice and records the product of the numbers [#permalink]

Show Tags

New post 12 Feb 2011, 04:45
1
This post received
KUDOS
10
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

76% (02:02) correct 24% (02:32) wrong based on 221 sessions

HideShow timer Statistics

Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt. What is the probability that the result of any attempt is an odd integer divisible by 25?

A. 7/216
B. 5/91
C. 13/88
D. 1/5
E. 3/8


Could anyone enlighten me on this tricky probability problem:

"If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an even number divisible by 25 at any attempt?"

My logic is that:
1) You need at least two fives in order to get 25. So: 1/6 x 1/6 x something
2) For the "something" above, you have three options: 2, 4, or 6, because the question asks for an even number. So: 1/6 x 1/6 x 1/2 (i.e. 3 favorable outcomes out of six) = 1/72

(Source: Veritas Prep, Combinatorics p.102)

However, the OA is 7/216, and I cannot fathom what my logic above is missing out. If anyone could lend a hand to solve this problem, I would appreciate.

Thanks!
[Reveal] Spoiler: OA

_________________

If you like my post, a kudos is not expected but appreciated


Last edited by Bunuel on 07 Jul 2013, 04:38, edited 1 time in total.
Renamed the topic and edited the question.

Kudos [?]: 44 [1], given: 8

Veritas Prep GMAT Discount CodesManhattan GMAT Discount CodesMagoosh Discount Codes
Intern
Intern
avatar
Joined: 06 Feb 2011
Posts: 27

Kudos [?]: 7 [0], given: 4

Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 12 Feb 2011, 05:57
Unfortunately, I am unable to get the answer 7/216 as well.

What I do say though is that I believe the answer could be 9/216.

Given 3 dice, the outcome that gives an even product divisible by 25 would be x,5,5 where x is an even number. And i believe x,5,5 is different from 5,x,5 or 5,5,x. Since there are 3 possible combinations of dice sequence, I get 9/216 by multiplying (1/2*1/6*1/6)*3.

But I don't get why the OA has to be 7/216 also..

Kudos [?]: 7 [0], given: 4

Manager
Manager
avatar
Status: Quant 50+?
Joined: 02 Feb 2011
Posts: 105

Kudos [?]: 30 [0], given: 22

Concentration: Strategy, Finance
Schools: Tuck '16, Darden '16
Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 12 Feb 2011, 09:45
I get 9/216 too.

1/6*1/6*1/2, but there are 3 combinations of this, the 1/2 can be in slot 1, 2 or 3.

(1/6*1/6*1/2)*3 = 9/216

Kudos [?]: 30 [0], given: 22

Manager
Manager
avatar
Joined: 16 Oct 2009
Posts: 206

Kudos [?]: 44 [0], given: 8

Schools: HEC Paris, , Tepper
Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 12 Feb 2011, 15:27
Thanks guys, for inspiration.

By the ways, (1/6 x 1/6/1/2)^3 = 1/216, instead of 9/216, no?
_________________

If you like my post, a kudos is not expected but appreciated

Kudos [?]: 44 [0], given: 8

Intern
Intern
avatar
Joined: 06 Feb 2011
Posts: 27

Kudos [?]: 7 [0], given: 4

Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 12 Feb 2011, 17:19
almostfamous wrote:
Thanks guys, for inspiration.

By the ways, (1/6 x 1/6/1/2)^3 = 1/216, instead of 9/216, no?


Nope, i am pretty sure its 9/216. Btw, does the OA contain explanations as well? Also mind sharing what the answer choices are since this should be a PS question

Kudos [?]: 7 [0], given: 4

Manager
Manager
avatar
Status: Quant 50+?
Joined: 02 Feb 2011
Posts: 105

Kudos [?]: 30 [0], given: 22

Concentration: Strategy, Finance
Schools: Tuck '16, Darden '16
Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 12 Feb 2011, 18:53
almostfamous wrote:
Thanks guys, for inspiration.

By the ways, (1/6 x 1/6/1/2)^3 = 1/216, instead of 9/216, no?




x3 not to the power of 3. There are 3 different ways it can happen.

Kudos [?]: 30 [0], given: 22

1 KUDOS received
Manager
Manager
avatar
Status: Quant 50+?
Joined: 02 Feb 2011
Posts: 105

Kudos [?]: 30 [1], given: 22

Concentration: Strategy, Finance
Schools: Tuck '16, Darden '16
Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 13 Feb 2011, 05:46
1
This post received
KUDOS
I have a theory. There is another question going similar to this but slightly different where:

Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as a result of the attempt. What is the probability that the result of any attempt is an odd integer divisible by 25?

In this situation the answer is 7/216 because 5x5x5 only occurs once, not 3 times.

Where as in our example 552 255 525 455 545 554 655 565 556 all occur (9 options)

Kudos [?]: 30 [1], given: 22

1 KUDOS received
Manager
Manager
User avatar
Joined: 17 Feb 2011
Posts: 188

Kudos [?]: 781 [1], given: 70

Concentration: Real Estate, Finance
Schools: MIT (Sloan) - Class of 2014
GMAT 1: 760 Q50 V44
Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 17 Feb 2011, 09:26
1
This post received
KUDOS
I also get 9/216: (1/6 x 1/6 x 3/6) x 3. The probabilities would be:

2 x 5 x 5
4 x 5 x 5
6 x 5 x 5
5 x 2 x 5
5 x 4 x 5
5 x 6 x 5
5 x 5 x 2
5 x 5 x 4
5 x 5 x 6

So, we have 216 possible outcomes and 9 desired results, as shown above. I strongly believe the OA is wrong.

Kudos [?]: 781 [1], given: 70

Current Student
avatar
Status: Up again.
Joined: 31 Oct 2010
Posts: 526

Kudos [?]: 560 [0], given: 75

Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42
Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 17 Feb 2011, 11:59
1
This post was
BOOKMARKED
almostfamous wrote:
Could anyone enlighten me on this tricky probability problem:

"If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an even number divisible by 25 at any attempt?"

My logic is that:
1) You need at least two fives in order to get 25. So: 1/6 x 1/6 x something
2) For the "something" above, you have three options: 2, 4, or 6, because the question asks for an even number. So: 1/6 x 1/6 x 1/2 (i.e. 3 favorable outcomes out of six) = 1/72

(Source: Veritas Prep, Combinatorics p.102)

However, the OA is 7/216, and I cannot fathom what my logic above is missing out. If anyone could lend a hand to solve this problem, I would appreciate.

Thanks!


Hi, the question on p102 is:

Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as a result of the attempt. What is the probability that the result of any attempt is an odd integer divisible by 25?

Answer is 7/216 and the explanation to me is quite convincing.

What is the source of this question?
_________________

My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

Kudos [?]: 560 [0], given: 75

Expert Post
6 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42630

Kudos [?]: 135779 [6], given: 12714

Re: Probability: even number divisible by 5 by throwing 3 dice [#permalink]

Show Tags

New post 17 Feb 2011, 13:23
6
This post received
KUDOS
Expert's post
11
This post was
BOOKMARKED
almostfamous wrote:
Could anyone enlighten me on this tricky probability problem:

"If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an even number divisible by 25 at any attempt?"

My logic is that:
1) You need at least two fives in order to get 25. So: 1/6 x 1/6 x something
2) For the "something" above, you have three options: 2, 4, or 6, because the question asks for an even number. So: 1/6 x 1/6 x 1/2 (i.e. 3 favorable outcomes out of six) = 1/72

(Source: Veritas Prep, Combinatorics p.102)

However, the OA is 7/216, and I cannot fathom what my logic above is missing out. If anyone could lend a hand to solve this problem, I would appreciate.

Thanks!


Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt. What is the probability that the result of any attempt is an ODD integer divisible by 25?
A. 7/216
B. 5/91
C. 13/88
D. 1/5
E. 3/8

In this case as 555 can occur only one way, then in nominator we'll have 3+3+1=7:
551 - 515 - 155;
553 - 535 - 355;
555.

\(P(55O)=\frac{7}{216}\) (if you want to calculate as in previous case then: \(P(55O)=P(551)+P(553)+P(555)=\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}+\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{7}{216}\))

Note that above 16 scenarios describe all cases of having at least two 5 on 3 dice: 7 cases give odd sum and 9 give even sum.

Answer: A.

If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an EVEN number divisible by 25 at any attempt?

\(P(55E)=\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{3}{6}=\frac{9}{216}\), we multiply by \(\frac{3!}{2!}\) or simply by 3 because 5-5-Even can occur in 3 different ways: 5-5-Even, 5-Even-5, or Even-5-5 (for example: 552, 525, or 255).

Check all 9 scenarios:
552 - 525 - 255;
554 - 545 - 455;
556 - 565 - 655.

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135779 [6], given: 12714

1 KUDOS received
Intern
Intern
avatar
B
Joined: 01 Sep 2014
Posts: 9

Kudos [?]: 2 [1], given: 1

Location: India
GMAT 1: 730 Q50 V37
GPA: 2.8
Re: Crowan throws 3 dice and records the product of the numbers [#permalink]

Show Tags

New post 01 Sep 2014, 20:57
1
This post received
KUDOS
7 favorable events (551, 553, 555, 515, 535, 155 & 355) out of 216 events. As only odd products of 25 are asked for.

Kudos [?]: 2 [1], given: 1

Intern
Intern
avatar
Joined: 02 Oct 2015
Posts: 14

Kudos [?]: 11 [0], given: 0

Re: Crowan throws 3 dice and records the product of the numbers [#permalink]

Show Tags

New post 06 Oct 2015, 23:42
if i am in hurry ..i will mark A as answer as Denominator has to a factor of 216 so option A and E are only valid ones..
in three throws 5 have to occur twice so permutation will be 3!/2! x 3(for three odd numbers 1,3,5) - 2(as for the case when 5 will appear in all three throws)=7 so 7/216

Kudos [?]: 11 [0], given: 0

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 14816

Kudos [?]: 288 [0], given: 0

Premium Member
Re: Crowan throws 3 dice and records the product of the numbers [#permalink]

Show Tags

New post 23 Nov 2017, 21:12
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

Kudos [?]: 288 [0], given: 0

Re: Crowan throws 3 dice and records the product of the numbers   [#permalink] 23 Nov 2017, 21:12
Display posts from previous: Sort by

Crowan throws 3 dice and records the product of the numbers

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


GMAT Club MBA Forum Home| About| Terms and Conditions| 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®.