Answer:
[tex]\dfrac{1}{4}[/tex]
Step-by-step explanation:
We are given that a card is drawn from a hat.
Probability that a red card is drawn, P(red) = [tex]\frac{1}{10}[/tex]
Probability that a blue card is drawn, P(blue) = [tex]\frac{1}{5}[/tex]
Probability that a black card is drawn, P(black) = [tex]\frac{1}{20}[/tex]
Probability that a pink card is drawn, P(pink) = [tex]\frac{1}{5}[/tex]
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:
[tex]P(A\cup B) = P(A) + P(B)[/tex]
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)
[tex]\Rightarrow \dfrac{1}{5}+\dfrac{1}{20}\\\Rightarrow \dfrac{4+1}{20}\\\Rightarrow \dfrac{5}{20}\\\Rightarrow \dfrac{1}{4}[/tex]
So, the probability that a blue or black card is drawn from the hat:
[tex]\dfrac{1}{4}[/tex]
