Answer:
[tex]\sf C. \quad \dfrac{1}{9}[/tex]
Step-by-step explanation:
Addition Law for Probability
[tex]\sf P(A \cup B)=P(A)+P(B)-P(A \cap B)[/tex]
Given:
[tex]\sf P(A)=\dfrac{1}{3}=\dfrac{3}{9}[/tex]
[tex]\sf P(B)=\dfrac{2}{9}[/tex]
[tex]\sf P(A \cup B)=\dfrac{4}{9}[/tex]
Substitute the given values into the formula and solve for P(A ∩ B):
[tex]\implies \sf P(A \cup B) = P(A)+P(B)-P(A \cap B)[/tex]
[tex]\implies \sf \dfrac{4}{9} = \sf \dfrac{3}{9}+\dfrac{2}{9}-P(A \cap B)[/tex]
[tex]\implies \sf P(A \cap B) = \sf \dfrac{3}{9}+\dfrac{2}{9}-\dfrac{4}{9}[/tex]
[tex]\implies \sf P(A \cap B) = \sf \dfrac{3+2-4}{9}[/tex]
[tex]\implies \sf P(A \cap B) = \sf \dfrac{1}{9}[/tex]
