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.htmlHope it helps.