Question

Let E and F be two independent events. The probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2 .Then,

• A. (a) P(E)=1/3, P(F)=1/4
  
• B. (b) P(E)=1/2, P(F)=1/6
  
• C. (c) P(E)=1/6, P(F)=1/2
  
• D. (d) P(E)=1/12, P(F)=1
  

Answer 1

The correct option is A

(a) P(E)=1/3, P(F)=1/4

We are given that
$P(E\cap F)=1/12andP(E^{\prime }\cap F^{\prime })=1/2$
As E and F are independent, we get
$P\left(E\right)P\left(F\right)=1/12\text{and}P\left(E^{\prime }\right)P\left(F^{\prime }\right)=1/2$
But $P\left(A^{\prime }\right)=1-P\left(A\right)$.
$\therefore (1-P(E\left)\right)(1-P(F\left)\right)=1/2$
$\Rightarrow 1-P\left(E\right)-P\left(F\right)+P\left(E\right)P\left(F\right)=1/2\Rightarrow P\left(E\right)+P\left(F\right)=7/12$
The quadratic equation whose roots are $P\left(E\right)$ and $P\left(F\right)$ is
$x^2-\left(P\right(E)+P(F\left)\right)x+P\left(E\right)P\left(F\right)=0\Rightarrow x^2-\frac{7}{12}x+\frac{1}{12}=0\Rightarrow 12x^2-7x+1=0\Rightarrow (3x-1)(4x-1)=0\Rightarrow x=1/3,1/4.$