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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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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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.

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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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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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

This one we can solve easily by bernoulli combinations.
3 ways for even sum: (1) EEEE + (2) OOOO + (3) 2E2O
P(1)=P(2)=0.5^4
P(3)=2C4x(0.5^2)x(0.5^2)
Sum = 1/2