Question 313596
P(K)=4/52=1/13
P(red)=13/52=1/4
P(red K)=2/52
Subtract the red Kings so as not to count them twice.
P(R or King)=P(K)+P(red)-P(red K)={{{4/52+13/52-2/52=15/52}}}
It's the same as all of the reds plus the 2 black kings.