Question

Please help!

Answer 1

$\\ \rm\leadsto P(A\cap B)=P(A)+P(B)-P(A\cup B)$

$\\ \rm\leadsto P(A\cap B)=\dfrac{1}{3}+\dfrac{2}{9}-\dfrac{4}{9}$

$\\ \rm\leadsto P(A\cap B)=\dfrac{3+2-4}{9}$

$\\ \rm\leadsto P(A\cap B)=\dfrac{1}{9}$

Answer 2

Answer:

$\sf C. \quad \dfrac{1}{9}$

Step-by-step explanation:

Addition Law for Probability

$\sf P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Given:

  $\sf P(A)=\dfrac{1}{3}=\dfrac{3}{9}$

  $\sf P(B)=\dfrac{2}{9}$

  $\sf P(A \cup B)=\dfrac{4}{9}$

Substitute the given values into the formula and solve for P(A ∩ B):

$\implies \sf P(A \cup B) = P(A)+P(B)-P(A \cap B)$

$\implies \sf \dfrac{4}{9} = \sf \dfrac{3}{9}+\dfrac{2}{9}-P(A \cap B)$

$\implies \sf P(A \cap B) = \sf \dfrac{3}{9}+\dfrac{2}{9}-\dfrac{4}{9}$

$\implies \sf P(A \cap B) = \sf \dfrac{3+2-4}{9}$

$\implies \sf P(A \cap B) = \sf \dfrac{1}{9}$