Of three independent events, the chance that only the first occurs is a, the chance that only the second occurs is b and the chance of only third is c. Show that the chances of three events are, respectively, a/(a+x), b/(b+x), c/(c+x) where x is a root of the equation $$(a+x)(b+x)(c+x)=x²$$
My attempt:
Let A, B, C be three independent events having probability p, q, and r, respectively. Then, we have, $$p(1-q)(1-r)=a$$ $$(1-p)q(1-r)=b$$ $$(1-p)(1-q)r=c$$
And after this I do not get how to relate my equations to x. I don't see anything to go on from here.
Set $x = \mathbb{P}(A^c \cap B^c \cap C^c)$. Then, using independence, $$ \mathbb{P}(A) = \frac{\mathbb{P}(A \cap B^c \cap C^c)}{\mathbb{P}(B^c \cap C^c)} = \frac{\mathbb{P}(A \cap B^c \cap C^c)}{\mathbb{P}(A \cap B^c \cap C^c) + \mathbb{P}(A^c \cap B^c \cap C^c) } = \frac{\mathbb{P}(A \cap B^c \cap C^c)}{\mathbb{P}(A \cap B^c \cap C^c) + x},$$ and this last quantity is of course $(\frac{a}{a+x})$. You can replicate this computation for the two other cases. Moreover, using again the independance,
$$ (a+x)(b+x)(c+x) = \mathbb{P}(B^c \cap C^c) \mathbb{P}(A^c \cap C^c) \mathbb{P}(B^c \cap A^c) = \mathbb{P}(A^c)^2 \mathbb{P}(B^c)^2 \mathbb{P}(C^c)^2 = \mathbb{P}(A^c \cap B^c \cap C^c)^2 = x^2. $$