Question

The odds in favor of standing first of three students appearing in an examination are $1:2,2:5$ and $1:7$ respectively. The probability that either of them will stand first, is

• A. $\frac{125}{168}$
• B. $\frac{75}{168}$
• C. $\frac{32}{168}$
• D. $\frac{4}{168}$

Answer 1

The correct option is A $\frac{125}{168}$

Let the three student be $P,Q$ and $R$
Let $A,B,C$ denote the events of stading first of the three students $P,Q,R$ respectively
Given, odds in favour of $A=1:2$
odds in favour of $B=2:5$
and odds in favour of $C=1:7$
$\therefore P\left(A\right)=\frac{1}{1+2}=\frac{1}{3},P\left(B\right)=\frac{2}{2+5}=\frac{2}{7},P\left(C\right)=\frac{1}{1+7}=\frac{1}{8}$

Since events $A,B,C$ are mutually exclusives
$\therefore P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$

$=\frac{1}{3}+\frac{2}{7}+\frac{1}{8}=\frac{56+48+21}{168}=\frac{125}{168}.$