Question

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.

• A. $\frac{5}{12}$
• B. $\frac{1}{3}$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

Answer: (D)

$\frac{4}{3}$

Let $P\left(E\right)=x,P\left(F\right)=y$

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

$\Rightarrow P\left(E\right)P\left(F\right)=\frac{1}{12}$

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

$P(E^{\prime }\cap F^{\prime })=\frac{1}{2}$

$\Rightarrow (1-P(E\left)\right)(1-P(F\left)\right)=\frac{1}{2}$

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

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

$x+y=1+xy-\frac{1}{2}=1+\frac{1}{12}-\frac{1}{2}=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$

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

$(x-y{)}^2=(x+y{)}^2-4xy$

$(x-y{)}^2=\frac{49}{144}-\frac{1}{3}$

$(x-y{)}^2=\frac{49-48}{144}=\frac{1}{144}$

$x-y=\frac{1}{12}$

$2x=\frac{8}{12}$

$\Rightarrow x=\frac{4}{12}$

$y=\frac{3}{12},x=\frac{4}{12}$

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