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horsena [70]
1 year ago
12

Please help!

Mathematics
2 answers:
Likurg_2 [28]1 year ago
7 0

\\ \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}

Mrrafil [7]1 year ago
4 0

Answer:

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

Step-by-step explanation:

<u>Addition Law for Probability</u>

\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}

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