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

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Gambling with 4 dice, what is the probability of getting an [#permalink] New post 24 May 2012, 01:31
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48% (02:33) correct 51% (01:14) wrong based on 2 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

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Re: Gambling with 4 dice, what is the probability of getting an [#permalink] New post 24 May 2012, 01:52
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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] 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.

Regards,
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink] New post 24 May 2012, 02:30
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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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. NEW!!!


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Re: Gambling with 4 dice, what is the probability of getting an [#permalink] 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] New post 27 Feb 2013, 05:06
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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.
Re: Gambling with 4 dice, what is the probability of getting an   [#permalink] 27 Feb 2013, 05:06
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