Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 01 Jul 2015, 04:58

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Gambling with 4 dice, what is the probability of getting an

Author Message
TAGS:
Senior Manager
Joined: 28 Mar 2012
Posts: 287
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 18

Kudos [?]: 216 [0], given: 23

Gambling with 4 dice, what is the probability of getting an [#permalink]  24 May 2012, 00:31
2
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

67% (02:32) correct 33% (01:22) wrong based on 79 sessions
Gambling with 4 dice, what is the probability of getting an even sum?

A. 3/4
B. 1/2
C. 2/3
D. 1/4
E. 1/3
[Reveal] Spoiler: OA

_________________

My posts: Solving Inequalities, Solving Simultaneous equations, Divisibility Rules

My story: 640 What a blunder!

Vocabulary resource: EdPrep

Math Expert
Joined: 02 Sep 2009
Posts: 28223
Followers: 4457

Kudos [?]: 44918 [5] , given: 6634

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]  24 May 2012, 00:52
5
KUDOS
Expert's post
1
This post was
BOOKMARKED
Gambling with 4 dice, what is the probability of getting an even sum?

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

Even sum can be obtained in following cases:
EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4;
OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4;
EEOO - $$\frac{4!}{2!2!}=6$$ cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

$$P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}$$.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 1/2.

Similar problem to practice: a-box-contains-100-balls-numbered-from-1-to-100-if-three-b-109279.html

Hope it helps.
_________________
Senior Manager
Joined: 28 Mar 2012
Posts: 287
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 18

Kudos [?]: 216 [0], given: 23

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]  24 May 2012, 01:26
Hi Bunuel,

The shortcut is perfect! Are you aware of any thread where we such approaches are discussed?

Similar solution can be applied to problem like no. of cases A occurs before B in a sequence, no. of cases where A & B are together
In both cases half of the total no. of cases would be the answer. Since, either A comes after B or before B, either A & B are together or separated.

Regards,
_________________

My posts: Solving Inequalities, Solving Simultaneous equations, Divisibility Rules

My story: 640 What a blunder!

Vocabulary resource: EdPrep

Math Expert
Joined: 02 Sep 2009
Posts: 28223
Followers: 4457

Kudos [?]: 44918 [0], given: 6634

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]  24 May 2012, 01:30
Expert's post
Hi Bunuel,

The shortcut is perfect! Are you aware of any thread where we such approaches are discussed?

Similar solution can be applied to problem like no. of cases A occurs before B in a sequence, no. of cases where A & B are together
In both cases half of the total no. of cases would be the answer. Since, either A comes after B or before B, either A & B are together or separated.

Regards,

DS questions on probability: search.php?search_id=tag&tag_id=33
PS questions on probability: search.php?search_id=tag&tag_id=54

Hope it helps.
_________________
Director
Joined: 08 Jun 2010
Posts: 644
Followers: 0

Kudos [?]: 45 [0], given: 114

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]  06 Feb 2013, 03:14
very hard, I want to follow this post. this is 51/51 level , I think
Director
Joined: 08 Jun 2010
Posts: 644
Followers: 0

Kudos [?]: 45 [1] , given: 114

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]  27 Feb 2013, 04:06
1
KUDOS
probability for EEEE to happen is

1/2*1/2*1/2*1/2=A

similarly probability for OOOO and OOEE, OEOE,...(there are 6 cases)

total 8 case

add 8 cases we have 8/A=1/2

is the result.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 5338
Followers: 310

Kudos [?]: 60 [0], given: 0

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]  01 May 2015, 02:02
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.
_________________
Director
Joined: 08 Jun 2010
Posts: 644
Followers: 0

Kudos [?]: 45 [0], given: 114

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]  01 May 2015, 02:02
Bunuel wrote:
Gambling with 4 dice, what is the probability of getting an even sum?

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

Even sum can be obtained in following cases:
EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4;
OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4;
EEOO - $$\frac{4!}{2!2!}=6$$ cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

$$P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}$$.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 1/2.

Similar problem to practice: a-box-contains-100-balls-numbered-from-1-to-100-if-three-b-109279.html

Hope it helps.

I can not say any word for this excellency
Re: Gambling with 4 dice, what is the probability of getting an   [#permalink] 01 May 2015, 02:02
Similar topics Replies Last post
Similar
Topics:
1 What is the probability that the sum of two dice will yield 2 08 Feb 2012, 13:50
10 Two dice are tossed once. The probability of getting an even 12 23 Jan 2012, 20:20
22 4 dices are thrown at the same time. What is the probability 24 27 Dec 2009, 22:21
6 fair dice are tossed. What is the probability that at 7 08 Oct 2007, 11:25
Two dice are rolled. What is the probability the sum will be 7 26 Jan 2006, 17:38
Display posts from previous: Sort by