By "standard number cube", I'm going to assume we're talking a 6-sided die.
We're asked for the #P(>3nnodd)#. To figure this out, we need to know #P(>3)# and #P(odd)# and then subtract out anything that meets both conditions because we'd be double adding - and that is written #P(>3uuodd)#, so the whole thing looks like this:
#P(>3nnodd)=P(>3)+P(odd)-P(>3uuodd)#
With #P(>3)# - that condition is met if we get a 4, 5, or 6, so that gives us #1/2#.
With #P(odd)# - that condition is met if we get a 1, 3, or 5, so that also gives us #1/2#.
With #P(>3uuodd)# - that condition is met only by the number 5, so that gives us #1/6#.
Substituting in:
#P(>3nnodd)=P(>3)+P(odd)-P(>3uuodd)#
#P(>3nnodd)=1/2+1/2-1/6=5/6#
