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:
http://gmatclub.com/forum/a-box-contain ... 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?