If A and B are two independent events such th | vedclass

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

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$0.3$

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(C) Given that $A$ and $B$ are independent events, $A$ and $B^c$ are also independent events.$P(B) = \frac{2}{7}$, so $P(B^c) = 1 - P(B) = 1 - \frac{2}{7} = \frac{5}{7}$.We are given $P(A \cup B^c) = 0.8$.Using the formula $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c)$ and the independence property $P(A \cap B^c) = P(A) \cdot P(B^c)$:$0.8 = P(A) + \frac{5}{7} - P(A) \cdot \frac{5}{7}$$0.8 - \frac{5}{7} = P(A) \cdot (1 - \frac{5}{7})$$\frac{5.6 - 5}{7} = P(A) \cdot \frac{2}{7}$$\frac{0.6}{7} = P(A) \cdot \frac{2}{7}$$P(A) = \frac{0.6}{2} = 0.3$.

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

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