Question

Three persons, Aman, Bipin and Mohan attempt a Mathematics problem independently. The odds in favour of Aman and Mohan solving the problem are $3:2$ and $4:1$ respectively and the odds against Bipin solving the problem are $2:1$. Find
(i) The probability that all the three will solve the problem.
(ii) The probability that problem will be solved.

Answer 1

Odd against $=\frac{1-p}{p}$
Odd in favour $=\frac{p}{1-p}$
$A$ : Event Aman solves problem
$B$ : Event Bipin solves problem
$C$ : Event Mohan solves problem

Aman
$\frac{3}{2}=\frac{p}{1-p}$
$\Rightarrow p=\frac{3}{5}$
$\Rightarrow P\left(A\right)=\frac{3}{5}$Mohan

$\frac{4}{1}=\frac{p}{1-p}$
$p=\frac{4}{5}$
$\Rightarrow P\left(C\right)=\frac{4}{5}$Bipin

$\frac{2}{1}=\frac{1-p}{p}$
$\Rightarrow p=\frac{1}{3}$
$\Rightarrow P\left(B\right)=\frac{1}{3}$

(i) $P(A\cap B\cap C)=P\left(A\right).P\left(B\right).P\left(C\right)$
$=\frac{3}{5}\times \frac{1}{3}\times \frac{4}{5}=\frac{4}{25}$

(ii) Probability that the problem is not solved = the probability that all the three fail to solve it.
$\therefore P(\overline{A}\cap \overline{b}\cap \overline{C})=P\left(\overline{A}\right).P\left(\overline{b}\right).P\left(\overline{C}\right)$$=(1-\frac{3}{5}).(1-\frac{1}{3}).(1-\frac{4}{5})$

$=\frac{2}{5}.\frac{2}{3}.\frac{1}{5}=\frac{4}{75}$

$\therefore$ The probability that the problem will be solved $=1-\frac{4}{75}=\frac{71}{75}$.