Discussion Forum

1)

Of the three independent events  $E_{1},E_{2} and E_{3}$  , the probability that only $E_{1}$ occurs is $\alpha$ , only $E_{2}$ occurs is β   and only $E_{3}$ occurs is $\gamma$ .Let  the probability p that none  of $E_{1}$ , $E_{2}$ or $E_{3}$ occurs satisfy the equations   $(\alpha-2\beta),p=\alpha\beta$ and   $ (\beta-3\gamma)$   $ p=2\beta\gamma$  .All the given probabilities are assumed to lie in the interval (0,1)

 Then, probability of occurrence of E₁/ probability of occurrence  of E₃ is equal to

Answer:

Option D

Explanation:

Concept involved

 For   the events to be independent.

 $P(E_{1}\cap E_{2}\cap E_{3})=P(E_{1}) P(E_{2})P(E_{3})$

$P(E_{1}\cap \overline{E}_{2}\cap \overline{E}_{3})=P$

                                    (on;y E₁ occurs)

 =   $P(E_{1}).(1-P(E_{2}))(1-P(E_{3}))$

 let x,y,z are probability of  $E_{1}$, $E_{2}$ and $E_{3}$ respectively

 $\therefore$             $\alpha$   = x(1-y)(1-z)   ......(i)

                                 $\beta$    = (1-x).y(1-z)         ......(ii)

                                  $\gamma$  = (1-x)(1-y) z     ..........(iii)

$\Rightarrow$   p = (1-x) (1-y) (1-z)        ..........(iv)

 Given   $(\alpha-2\beta)p=\alpha\beta$ 

  and     $(\beta-3\gamma)p=2\beta\gamma$    .......(v)

 From above equations

 x=2y  and y=3 z

  $\therefore$  x=6z   =  $\frac{x}{z}=6$