Question

Probability of occurrence of the event $B$ is

• A. $\frac{x}{a+b}$
• B. $\frac{(1-c)(x+b)}{ax}$
• C. $\frac{{(1-a)}^2+ab)}{1-a}$
• D. $\frac{(1-b)(a-c)}{ax}$

Answer 1

Answer: (B)

$\frac{(1-c)(x+b)}{ax}$

Given $x=P(\overline{A}\cdot \overline{B}\cdot \overline{C})=P\left[\right(\overline{A\times B})\cdot C]=P\left(C\right)-P\left[\right(A\times B)\cdot C]$
$=P\left(C\right)-P[A\cdot C\times B\cdot C]$
$=P\left(C\right)-P(A\cdot C)-P(B\cdot C)+P(A\cdot B\cdot C)$
$=P\left(C\right)-P\left(A\right)\cdot P\left(C\right)-P\left(B\right)\cdot P\left(C\right)-P\left(A\right)\cdot P\left(B\right)\cdot P\left(C\right)$

$=P\left(C\right)-[1-P(A\left)\right]-P\left(B\right)P\left(C\right)[1-P(A\left)\right]$
$=(1-P(A\left)\right)\left[P\right(C)-P(B\left)P\right(C\left)\right]$
$=P\left(C\right)\{1-P(A\left)\right\}\{1-P(B\left)\right\}$

$\Rightarrow x=P\left(C\right)(1-a)\{1-P(B\left)\right\}$
$\Rightarrow P\left(C\right)\{1-P(B\left)\right\}=\frac{x}{1-a}......i$

$\Rightarrow P\left(C\right)-P\left(C\right)P\left(B\right)=\frac{x}{1-a}$ and $P(A\cdot B\cdot C)=1-c$ $\Rightarrow P\left(A\right)P\left(B\right)P\left(C\right)=1-c$ $\Rightarrow P\left(B\right)P\left(C\right)=\frac{1-c}{a}......\left(ii\right)$ Also $P(A\cdot B\cdot C)=b$ from result of question (i) $\Rightarrow P\left(\overline{A}\right)P\left(\overline{B}\right)P\left(\overline{C}\right)=b$

$\Rightarrow \{1-P(A\left)\right\}\cdot \{1-P(B\left)\right\}\cdot \{1-P(C\left)\right\}=b$
$\Rightarrow \{1-P(B\left)\right\}\{1-P(C\left)\right\}=\frac{b}{1-a}....\left(iii\right)$

$\Rightarrow \frac{x\{1-P(C\left)\right\}}{(1-a)P\left(C\right)}=\frac{b}{1-a}$ by using (i)

$\Rightarrow \frac{P\left(C\right)}{1-P\left(C\right)}=\frac{x}{b}\Rightarrow P\left(C\right)=\frac{x}{x+b}$ Similarly

$P\left(B\right)=\frac{(1-c)(x+b)}{ax}$