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

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

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

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cyberjadugar 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

Looking for a one-liner answer...


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.

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

Answer: B.

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

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New post 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.

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

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cyberjadugar wrote:
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,


Search the following links.
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.
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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

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

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New post 06 Feb 2013, 04:14
very hard, I want to follow this post. this is 51/51 level , I think
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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

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New post 01 May 2015, 03:02
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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New post 01 May 2015, 03:02
Bunuel wrote:
cyberjadugar 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

Looking for a one-liner answer...


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.

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

Answer: B.

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.



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

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New post 24 Jun 2016, 15:30
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).

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

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

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

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New post 01 Jun 2017, 00:11
cyberjadugar 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

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

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New post 09 Jun 2017, 08:13
Bunuel wrote:
cyberjadugar 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

Looking for a one-liner answer...


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.

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

Answer: B.

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

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New post 10 Jun 2017, 09:33
SOUMYAJIT_ wrote:
Bunuel wrote:
cyberjadugar 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

Looking for a one-liner answer...


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.

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

Answer: B.

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.
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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

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

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New post 25 Jun 2017, 11:22
cyberjadugar 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


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
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Re: Gambling with 4 dice, what is the probability of getting an   [#permalink] 25 Jun 2017, 11:22
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