%%

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

JEE · Math · previous-year question

  1. A.$${4 \over 3}$$correct
  2. B.$${3 \over 2}$$
  3. C.$${1 \over 3}$$
  4. D.$${5 \over 12}$$

Answer

A. $${4 \over 3}$$

Explanation

Let P(E) = x and P(F) = y Now, $$P(E \cap F) = {1 \over {12}}$$ $$ \Rightarrow P(E)P(F) = {1 \over {12}}$$ $$ \Rightarrow xy = {1 \over {12}}$$ Also, $$P(E' \cap F') = {1 \over 2}$$ $$ \Rightarrow (1 - P(E))(1 - P(F)) = {1 \over 2}$$ $$ \Rightarrow (1 - x)(1 - y) = {1 \over 2}$$ $$ \Rightarrow 1 - x - y + xy = {1 \over 2}$$ $$ \Rightarrow x + y = 1 + xy - {1 \over 2}$$ $$ \Rightarrow x + y = 1 + {1 \over {12}} - {1 \over 2}$$ $$ \Rightarrow x + y = {1 \over 2} + {1 \over {12}} = {7 \over {12}}$$ ........ (1) Now, $${(x - y)^2} = {(x + y)^2} - 4xy$$ $$ \Rightarrow {(x - y)^2} = {{49} \over {144}} - {1 \over 3} \Rightarrow {{49 - 98} \over {144}} \Rightarrow {1 \over {144}}$$ $$ \Rightarrow (x - y) = {1 \over {12}} \Rightarrow x - y = {1 \over 2}$$ ........ (2) From Eqs. (1) and (2), we get $$(x + y)(x - y) = {7 \over {12}} + {1 \over {12}}$$ $$ \Rightarrow x = {4 \over {12}};y = {3 \over {12}}$$ $$ \Rightarrow {x \over y} = {4 \over 3} = {{P(E)} \over {P(F)}}$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions