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

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Aicharya · Accepted Answer

$\because P(B)=\frac{2}{7}$ and $P\left(A \cup B^c\right)=0.8$ $P\left(B^c\right)=1-\frac{2}{7}=\frac{5}{7}$ We know that $P\left(A \cup B^c\right)=P(A)+P\left(B^c\right)-P(A) \cdot P\left(B^c\right)$ $\Rightarrow \quad 0.8=P(A)+\frac{5}{7}-\frac{5}{7} P(A)$ $0.8=\frac{5}{7}+\frac{2}{7} P(A)$ $\Rightarrow \quad 5.6-5=2 P(A)$ $\Rightarrow \quad P(A)=0.3$

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

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