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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

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.

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

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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.

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.

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

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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).

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

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01 May 2015, 02:02

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.

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

Show Tags

24 Jun 2016, 14:30

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).

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.
_________________

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