Let E and F be two independent events. The&nb | vedclass

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Question

$P\left( {E \cap F} \right) = P(E) \cdot P(F) = \frac{1}{{12}}$

$P\left( {\bar E \cap \bar F} \right) = P(\bar E) \cdot P(\bar F) = \frac{1}{2}$

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

$P\left( {E \cap F} \right) = P(E) \cdot P(F) = \frac{1}{{12}}$

$P\left( {\bar E \cap \bar F} \right) = P(\bar E) \cdot P(\bar F) = \frac{1}{2}$

$(1-P(E))(1-P(F))=\frac{1}{2}$

Let $P(E)=x$

$P(F) =y $

$=1-x-y+x y=\frac{1}{2}$

$1-x-y=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$

$\boxed{x + y = \frac{7}{{12}}}$

$x+\frac{1}{12 x}=\frac{7}{12}$

$\left[\because x \cdot y=\frac{1}{12}\right]$

$12 x^{2}-7 x+1=0$

$12 x^{2}-4 x-3 x+1=0$

$(4 x-1)(3 x-1)=0$

$x=\frac{1}{3}, x=\frac{1}{4}$

$y=\frac{1}{4}, y=\frac{1}{3}$

$\frac{x}{y}=\frac{1 / 3}{1 / 4}=\frac{4}{3}$ or $\frac{1 / 4}{1 / 3}=\frac{3}{4}$

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happen is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2}$ , then a value of $\frac{{P(E)}}{{P\left( F \right)}}$ is

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