Question

For a post three persons 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 odds in favour of B to be selected is:

• A. 3 : 7
• B. 7 : 3
• C. 1 : 4
• D. 4 : 1

Answer 1

The correct option is A 3 : 7

Let $E_1,E_2,E_3$ be the events of selection of A, B, C respectively.
Let $P\left(E_3\right)=x$, then $P\left(E_2\right)=3X,P\left(E_1\right)=6x$
Since $E_1,E_2,E_3$ are mutually exclusive and exhaustive events,
$P(E_1\cup E_2\cup E_3)=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1\Rightarrow 10x=1\Rightarrow x=\frac{1}{10}\therefore P\left(E_2\right)=\frac{3}{10}$.
Hence, Odds in favour of B = 3 : 7