Let E and F be two independent events.

Question

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 :

Accepted Answer

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

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

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

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

Solutions · 1