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 $\left(3p_1+6p_2+10p_3\right)$.

Answer 1

Answer: 6

$p_1\left(1-p_2\right)\left(1-p_3\right)=\frac{1}{4}$ ....(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)
$(1)÷(2)$
$\frac{p_1}{p_2}\left(\frac{1-p_2}{1-p_1}\right)=2\Rightarrow \frac{2\left(1-p_1\right)}{p_1}=\left(\frac{1}{p_2}-1\right)\Rightarrow \frac{2\left(1-p_1\right)+p_1}{p_1}=\frac{1}{p_2}\Rightarrow \frac{2-p_1}{p_1}=\frac{1}{p_2}$
$\Rightarrow p_2=\frac{p_1}{2-p_1}$
$(1)÷(3)$
$\frac{p_1\left(1-p_3\right)}{p_3\left(1-p_1\right)}=3\Rightarrow \frac{1-p_3}{p_3}=\frac{3\left(1-p_1\right)}{p_1}\Rightarrow \frac{1}{p_3}=\frac{3\left(1-p_1\right)}{p_1}+1=\frac{3-2p_1}{p_1}\Rightarrow p_3=\frac{p_1}{3-2p_1}$
Now substituting $p_2$ and $p_3$ in (1) solving the cubic in $p_1$ we get
$p_1=\frac{1}{2},p_2=\frac{1}{3};p_3=\frac{1}{4}$
$\frac{3}{2}+2+\frac{5}{2}=6$