I am not able to solve this particular problem..I tried with some algebraic methods but I am not able to arrive at answers.

Question

Of three independent events, the probability that the first only should happens is 1/4, the probability that the second only should happen is 1/8, and the probability that the third only should happen is 1/12. Obtain the unconditional probabilities of the three events.

Answer 1

$p1(1-p2)(1-p3) = 1/4, (1-p1)p2(1-p3)=1/8, (1 -p1)(1-p2)p3 = 1/12 $.

Answer 2

Let $x,y,z$ be three unknown probabilities. There are two series of solutions of three equations $$ \begin{cases}x(1-y)(1-z)=\frac14 \tag{1}\cr (1-x)y(1-z)=\frac18 \cr (1-x)(1-y)z=\frac{1}{12} \end{cases} $$ $x=\frac12$, $y=\frac13$, $z=\frac14$ and $x=\frac{19-\sqrt{73}}{24}\approx 0.4356665$, $y=\frac{13-\sqrt{73}}{16}\approx 0.27849977$, $z=\frac{11-\sqrt{73}}{12}\approx 0.2046664$. Both solutions are valid probabilities of these independent events.

To solve these equations, one can divide the first by the second, then first by the third and get $$ \frac{x(1-y)}{(1-x)y} = 2, \quad \frac{x(1-z)}{(1-x)z} = 3. $$ If we denote $\frac{1-z}{z}=t>0$, then we can express all variables as a functions of $t$: $$ x=\frac{3}{t+3}, \quad y=\frac{3}{2t+3}, \quad z=\frac1{1+t}. $$ Substitute these exprssions into any of equations (1) (say, first), get cubic equation with respect to $t$: $$ 2t^3-13t^2+18t+9=0. $$ We can try to find root check all the divisors of the free term of the equation. $t=3$ is a root. After that we can rewrite $2t^3-13t^2+18t+9 = (t-3)(2t^2-7t-3)$ and find two more roots: $t=\frac{7+\sqrt{73}}{4}$, $t=\frac{7-\sqrt{73}}{4}<0$.

Finally, we have two roots: $t=3$ and $t=\frac{7+\sqrt{73}}{4}$ which give the probabilities above.