Answer:

Step-by-step explanation:
We are given that a card is drawn from a hat.
Probability that a red card is drawn, P(red) = 
Probability that a blue card is drawn, P(blue) = 
Probability that a black card is drawn, P(black) = 
Probability that a pink card is drawn, P(pink) = 
All these events are mutually exclusive i.e. they do not have anything in common.
For any two events A and B which are mutually exclusive, with probability P(A) and P(B), the probability that any one of the two events occur:

Here event A can be thought of as drawing a blue card and
event B can be thought of as drawing a black card
So, P(blue or black) = P(blue) + P(black)

So, the probability that a blue or black card is drawn from the hat:
