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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 $\frac{P\left (E \right )}{P\left ( F \right )}$ is

Option 1)
$\frac{4}{3}$

Option 2)
$\frac{3}{2}$

Option 3)
$\frac{1}{3}$

Option 4)
$\frac{5}{12}$

Answers (1)

As we learnt in

Independent events -

If A and B are independent events then probability of occurrence of A is not affected by occurrence or non occurrence of event B.

$\therefore P\left ( \frac{A}{B} \right )= P\left ( A \right )$

and $\dpi{100} \therefore P\left ( A\cap B \right )= P\left ( B \right )\cdot P\left ( \frac{A}{B} \right )$

so $\therefore P\left ( A\cap B \right )= P\left ( A \right )\cdot P \left ( B \right )= P\left ( AB \right )$

-

$P(E \cap F)=P(E).P(F)=\frac{1}{12}$

But $P(E)=x;\ \;P(F)=y$

$xy=\frac{1}{12}$

and $xy=(1-x)\: (1-y)=\frac{1}{2}$

$1-(x+y)+xy=\frac{1}{2}$

$x+y=\frac{7}{12}$

If we make a quadratic equation with x,y as roots

$m^{2}-\frac{7}{12}m+\frac{1}{12}=0$

$m=\frac{4}{3}\ \; \; \;or\ \; \frac{3}{4}$

$so\ \;\frac{x}{y}=\frac{4}{3}$

Option 1)

$\frac{4}{3}$

This option is correct

Option 2)

$\frac{3}{2}$

This option is incorrect

Option 3)

$\frac{1}{3}$

This option is incorrect

Option 4)

$\frac{5}{12}$

This option is incorrect

Posted by

Sabhrant Ambastha

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