Question

Of the 3 independent events A, B and C the probability that the event 'A' only should happen is $\frac{1}{4}$, the event ' $B$ ' only should happen is $1 / 8$ \& $C$ only should happen is $\frac{1}{12}$. If the unconditional probabilities of 3 events, (given that they have rational values) are $p _1, p _2$ and $p _3$ respectively, find the value of $\left(3 p _1+6 p _2+10 p _3\right)$.

Answer 1

$p_1\left(1-p_2\right)\left(1-p_3\right)=\frac{1}{4}$ ....(1)
$p _2\left(1- p _1\right)\left(1- p _3\right)=\frac{1}{8}$ ....(2)
$p_3\left(1-p_2\right)\left(1-p_1\right)=\frac{1}{12} $ ....(3)
$(1) \div(2) $
$\frac{p_1}{p_2}\left(\frac{1-p_2}{1-p_1}\right)=2 \Rightarrow \frac{2\left(1-p_1\right)}{p_1}=\left(\frac{1}{p_2}-1\right) \Rightarrow \frac{2\left(1-p_1\right)+p_1}{p_1}=\frac{1}{p_2} \Rightarrow \frac{2-p_1}{p_1}=\frac{1}{p_2} $
$\Rightarrow \quad p_2=\frac{p_1}{2-p_1}$
$(1) \div(3)$
$\frac{p_1\left(1-p_3\right)}{p_3\left(1-p_1\right)}=3 \Rightarrow \frac{1-p_3}{p_3}=\frac{3\left(1-p_1\right)}{p_1} \Rightarrow \frac{1}{p_3}=\frac{3\left(1-p_1\right)}{p_1}+1=\frac{3-2 p_1}{p_1} \Rightarrow p_3=\frac{p_1}{3-2 p_1}$
Now substituting $p_2$ and $p_3$ in (1) solving the cubic in $p_1$ we get
$p _1=\frac{1}{2}, p _2=\frac{1}{3} ; p _3=\frac{1}{4} $
$\frac{3}{2}+2+\frac{5}{2}=6 $