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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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24 May 2012, 01:31
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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
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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 oneliner 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: aboxcontains100ballsnumberedfrom1to100ifthreeb109279.htmlHope it helps.
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Kellogg MMM ThreadMaster
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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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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06 Feb 2013, 04:14
very hard, I want to follow this post. this is 51/51 level , I think
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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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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 oneliner 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: aboxcontains100ballsnumberedfrom1to100ifthreeb109279.htmlHope it helps. I can not say any word for this excellency
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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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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Gambling with 4 dice, what is the probability of getting an [#permalink]
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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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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 oneliner 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/aboxcontain ... 09279.htmlHope 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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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 oneliner 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/aboxcontain ... 09279.htmlHope 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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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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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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