Question

For a post, three-person, $A,B$ and $C$ appear in the interview. The probability of $A$ being selected is twice that of $B$ and the probability of $B$ being selected is thrice that of $C$, then the individual probabilities of $A,B,C$ respectively are:

• A. $P\left(A\right)=\frac{1}{10},P\left(B\right)=\frac{3}{10},P\left(C\right)=\frac{6}{10}$
  
• B. $P\left(A\right)=\frac{6}{10},P\left(B\right)=\frac{3}{10},P\left(C\right)=\frac{1}{10}$
• C. $P\left(A\right)=\frac{3}{10},P\left(B\right)=\frac{1}{10},P\left(C\right)=\frac{6}{10}$
• D. none of the above

Answer 1

Answer: (B)

$P\left(A\right)=\frac{6}{10},P\left(B\right)=\frac{3}{10},P\left(C\right)=\frac{1}{10}$

Let $A_1,A_{2}and{A}_3$ be three events, as defined below:
$A_1=$ Person A is selected,
$A_2=$ Person B is selected,
$A_3$= Person C is selected.
We have,
$P\left(A_1\right)=2P\left(A_2\right)$ and $P\left(A_2\right)=2P\left(A{)}_3\right)$
$\Rightarrow P\left(A_1\right)=6P\left(A_3\right)$ and $P\left(A_2\right)=3P\left(A_3\right)$
Since $A_1,A_2,A_3$ are mutually exclusive and exhaustive events.
$\therefore A_1\cup A_2\cup A_3=S$
$\Rightarrow P(A_1\cup A_2\cup A_3)=P\left(S\right)$
$\Rightarrow P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right)=1$
[Since $A_1,A_2,A_3$ are mutually exclusive]
$\Rightarrow 6P\left(A_3\right)+3P\left(A_3\right)+P\left(A_3\right)=1$
$\Rightarrow 10P\left(A_3\right)=1$
$\Rightarrow P\left(A_3\right)=\frac{1}{10}$
$\therefore P\left(A_1\right)=\frac{6}{10}$ and $P\left(A_2\right)=\frac{3}{10}$