QJEEM17Maf3ea: JEE Main 2017 Mathematics - Class 12 JEEM17Maf3ea Solution: Let E and F be two independent events. The probability that…

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

$(x + y)(x - y) = {7 \over {12}} + {1 \over {12}}$

$\Rightarrow x = {4 \over {12}};y = {3 \over {12}}$

$\Rightarrow {x \over y} = {4 \over 3} = {{P(E)} \over {P(F)}}$