It is currently 22 Nov 2017, 19:21

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:

Hide Tags

Joined: 29 Mar 2012
Posts: 320

Kudos [?]: 528 [1], given: 23

Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

24 May 2012, 01:31
1
KUDOS
14
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

59% (01:28) correct 41% (01:47) wrong based on 300 sessions

HideShow timer Statistics

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

Kudos [?]: 528 [1], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133076 [6], given: 12403

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

24 May 2012, 01:52
6
KUDOS
Expert's post
7
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.
_________________

Kudos [?]: 133076 [6], given: 12403

Joined: 29 Mar 2012
Posts: 320

Kudos [?]: 528 [0], given: 23

Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

24 May 2012, 02:26
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,

Kudos [?]: 528 [0], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133076 [0], given: 12403

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

24 May 2012, 02:30
Expert's post
1
This post was
BOOKMARKED
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.
_________________

Kudos [?]: 133076 [0], given: 12403

VP
Joined: 09 Jun 2010
Posts: 1393

Kudos [?]: 168 [0], given: 916

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

06 Feb 2013, 04:14
very hard, I want to follow this post. this is 51/51 level , I think
_________________

visit my facebook to help me.
on facebook, my name is: thang thang thang

Kudos [?]: 168 [0], given: 916

VP
Joined: 09 Jun 2010
Posts: 1393

Kudos [?]: 168 [1], given: 916

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

27 Feb 2013, 05:06
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.
_________________

visit my facebook to help me.
on facebook, my name is: thang thang thang

Kudos [?]: 168 [1], given: 916

VP
Joined: 09 Jun 2010
Posts: 1393

Kudos [?]: 168 [0], given: 916

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

01 May 2015, 03:02
1
This post was
BOOKMARKED
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
_________________

visit my facebook to help me.
on facebook, my name is: thang thang thang

Kudos [?]: 168 [0], given: 916

Intern
Joined: 05 Apr 2016
Posts: 36

Kudos [?]: 1 [0], given: 23

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

06 Jul 2016, 11:48
I got it wrong. Why am I wrong?
e+e=e
o+o=e
e+o=o

So it's even in 2/3 of the times.

Kudos [?]: 1 [0], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133076 [0], given: 12403

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

07 Jul 2016, 04:38
Mariwa wrote:
I got it wrong. Why am I wrong?
e+e=e
o+o=e
e+o=o

So it's even in 2/3 of the times.

_________________

Kudos [?]: 133076 [0], given: 12403

Director
Joined: 17 Dec 2012
Posts: 623

Kudos [?]: 535 [0], given: 16

Location: India
Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

01 Jun 2017, 00:11
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

1.We can throw either an even or an odd with a die and so there are 16 possibilities .
2. No even and 4 odd, the sum of 4 odds being even and occurring once
3. 1 even and 3 odd, the sum being odd and occurring 4 times
4. 2 even and 2 odd the sum being even and occurring 6 times
5. 3 even and 1 odd the sum being odd and occurring 4 times
6. 4 even and no odd the sum being even and occurring once
7. 8 times , the sum is even and 8 times the sum is odd for a probability of 1/2.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/regularcourse.php

Standardized Approaches

Kudos [?]: 535 [0], given: 16

Intern
Joined: 08 Mar 2016
Posts: 38

Kudos [?]: 5 [0], given: 13

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

09 Jun 2017, 08:13
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: http://gmatclub.com/forum/a-box-contain ... 09279.html

Hope it helps.

Brunnel, can you help me understand how you got 4!/2!2! for EEOO cases? I understand to arrange we need to do 4!, but can you break the 2!2! part here? I intutively understand its because of the pair of O and E, but can you explain it further?

Kudos [?]: 5 [0], given: 13

Math Expert
Joined: 02 Sep 2009
Posts: 42305

Kudos [?]: 133076 [0], given: 12403

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

10 Jun 2017, 09:33
SOUMYAJIT_ wrote:
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: http://gmatclub.com/forum/a-box-contain ... 09279.html

Hope it helps.

Brunnel, can you help me understand how you got 4!/2!2! for EEOO cases? I understand to arrange we need to do 4!, but can you break the 2!2! part here? I intutively understand its because of the pair of O and E, but can you explain it further?

THEORY:

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:

$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, the number of arrangements of four letters EEOO is 4!/(2!2!).

Hope it's clear.
_________________

Kudos [?]: 133076 [0], given: 12403

BSchool Forum Moderator
Joined: 17 Jun 2016
Posts: 449

Kudos [?]: 215 [0], given: 199

Location: India
GMAT 1: 720 Q49 V39
GMAT 2: 710 Q50 V37
GPA: 3.65
WE: Engineering (Energy and Utilities)
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

Show Tags

25 Jun 2017, 11:22
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

What can a sum of two positive integers (or for that matter any integers) be ?
It can only be EVEN or ODD..
there is no other possibility ...(note that I said INTEGERS - not just any numbers)

So, a dice when rolled, can only produce integers ..
So summing the results of any number of dice can only produce an ODD or an EVEN output.

Hence, there are 50% chance that our output will be EVEN (and 50% chances that output will be ODD)
So, Option B 1/2
_________________

Kudos [?]: 215 [0], given: 199

Senior Manager
Joined: 03 Apr 2013
Posts: 277

Kudos [?]: 44 [0], given: 856

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

Show Tags

03 Jul 2017, 01:51
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: http://gmatclub.com/forum/a-box-contain ... 09279.html

Hope it helps.

Hi Bunuel, Please help me with the concept that you have used in the second solution. You say that the probability will be 1/2 because the sum will either be even or be odd. 1/2 means that each of these events are equally likely. While solving Combinations problems, I have seen a similar approach from you in some questions. My question is, how can one infer whether the cases possible are equally likely? This will help save valuable time in exam once understood. Thank you for your help
_________________

Spread some love..Like = +1 Kudos

Kudos [?]: 44 [0], given: 856

Gambling with 4 dice, what is the probability of getting an   [#permalink] 03 Jul 2017, 01:51
Display posts from previous: Sort by