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(C) Given that $A$ and $B$ are independent events, $P(A \cap B) = P(A) \cdot P(B)$.We are given $P(B) = \frac{2}{7}$, so $P(B^c) = 1 - \frac{2}{7} = \

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$\frac{1}{2}$

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(C) Given that $A$ and $B$ are independent events, $P(A \cap B) = P(A) \cdot P(B)$.We are given $P(B) = \frac{2}{7}$, so $P(B^c) = 1 - \frac{2}{7} = \frac{5}{7}$.We are given $P(A \cup B^c) = 0.8$.Using the formula $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) = 0.8$.Since $A$ and $B$ are independent, $A$ and $B^c$ are also independent, so $P(A \cap B^c) = P(A) \cdot P(B^c)$.Substituting this into the equation: $P(A) + P(B^c) - P(A) \cdot P(B^c) = 0.8$.$P(A)(1 - P(B^c)) = 0.8 - P(B^c)$.$P(A)(1 - \frac{5}{7}) = 0.8 - \frac{5}{7}$.$P(A)(\frac{2}{7}) = \frac{4}{5} - \frac{5}{7} = \frac{28 - 25}{35} = \frac{3}{35}$.$P(A) = \frac{3}{35} \cdot \frac{7}{2} = \frac{3}{10} = 0.3$.Now, $P(A \cap B) = P(A) \cdot P(B) = \frac{3}{10} \cdot \frac{2}{7} = \frac{6}{70} = \frac{3}{35}$.Finally, $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{10} + \frac{2}{7} - \frac{3}{35} = \frac{21 + 20 - 6}{70} = \frac{35}{70} = \frac{1}{2}$.

If $A$ and $B$ are two independent events such that $P(B)=\frac{2}{7}$ and $P\left(A \cup B^c\right)=0.8$, then $P(A \cup B)$ $=$

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