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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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