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A and $\mathrm{B}$ are independent events. The probability that both A and $\mathrm{B}$ occur is $\frac{1}{20}$ and the probability that neither of them occurs is $\frac{3}{5}$. The probability of occurrence of $\mathrm{A}$ is
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$\frac{1}{4}$, $\frac{1}{5}$
\(P(A \cap B)=P(A) P(B)=\frac{1}{20} \Rightarrow P(B)=\frac{1}{2 O P(A)}\)
\(P(\bar{A} \cap \bar{B})=\frac{3}{5}=1-P(A \cup B)\)
\(\Rightarrow \frac{3}{5}=1-P(A)-P(B)+P(A \cap B)\)
\(\Rightarrow \frac{3}{5}=1-P(A)-\frac{1}{20 P(A)}+\frac{1}{20}\)
\(\Rightarrow \frac{3}{5}=\frac{21}{20}-P(A)-\frac{1}{20 P(A)}\)
\(\Rightarrow \frac{12-21}{20}=-P(A)-\frac{1}{20 P(A)}\)
Let \(P(A)=x\)
\(\Rightarrow \frac{-9}{20}=\frac{-20 x^2-1}{20 x}\)
\(\Rightarrow 20 x^2-9 x+1=0\)
\(\Rightarrow(4 x-1)(5 x-1)=0\)
\(\Rightarrow x=\frac{1}{4} \frac{1}{5}\)
i. e. \(P(A)=\frac{1}{4} P(A)=\frac{1}{5}\)
\(P(\bar{A} \cap \bar{B})=\frac{3}{5}=1-P(A \cup B)\)
\(\Rightarrow \frac{3}{5}=1-P(A)-P(B)+P(A \cap B)\)
\(\Rightarrow \frac{3}{5}=1-P(A)-\frac{1}{20 P(A)}+\frac{1}{20}\)
\(\Rightarrow \frac{3}{5}=\frac{21}{20}-P(A)-\frac{1}{20 P(A)}\)
\(\Rightarrow \frac{12-21}{20}=-P(A)-\frac{1}{20 P(A)}\)
Let \(P(A)=x\)
\(\Rightarrow \frac{-9}{20}=\frac{-20 x^2-1}{20 x}\)
\(\Rightarrow 20 x^2-9 x+1=0\)
\(\Rightarrow(4 x-1)(5 x-1)=0\)
\(\Rightarrow x=\frac{1}{4} \frac{1}{5}\)
i. e. \(P(A)=\frac{1}{4} P(A)=\frac{1}{5}\)
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