Question
Of three independent events $E_{1},E_{2},E_{3}$, the chance that only $E_{1}$ occurs is a , that only $E_{2}$ occurs is b , and only $E_{3}$ occurs is c and probability of none of $E_{1},E_{2},E_{3}$ occuring is x . Then
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Solution
Sol. [Key Concept: Three independent events: set up the pairwise/triple intersection probabilities and solve.]
Let $\mathbf{P}\left( {\bar{E}}_{1} \cap {\bar{E}}_{2} \cap {\bar{E}}_{3} \right) = \mathbf{x} = \mathbf{P}\left( {\bar{E}}_{1} \right)P\left( {\bar{E}}_{2} \right)P\left( {\bar{E}}_{3} \right)$
$$\begin{matrix} & P\left( E_{1} \cap {\bar{E}}_{2} \cap {\bar{E}}_{3} \right) = a = P\left( E_{1} \right)P\left( {\bar{E}}_{2} \right)P\left( {\bar{E}}_{3} \right) \\ P\left( {\bar{E}}_{1} \cap E_{2} \cap {\bar{E}}_{3} \right) = b & \ = P\left( {\bar{E}}_{1} \right)P\left( E_{2} \right)P\left( {\bar{E}}_{3} \right) \\ P\left( {\bar{E}}_{1} \cap {\bar{E}}_{2} \cap E_{3} \right) & \ = x = P\left( {\bar{E}}_{1} \right)P\left( {\bar{E}}_{2} \right)P\left( E_{3} \right) \\ \frac{a}{x} & \ = \frac{P\left( E_{1} \right)}{P\left( {\bar{E}}_{1} \right)} = \frac{P\left( E_{1} \right)}{1 - P\left( E_{1} \right)} \Rightarrow P\left( E_{1} \right) = \frac{a}{a + x} \\ P\left( {\bar{E}}_{1} \right)P\left( {\bar{E}}_{2} \right)P\left( {\bar{E}}_{2} \right) & \ = x = \frac{x}{a + x}\frac{x}{b + x}\frac{x}{c + x} \\ \Rightarrow (a + x)(b + x)(c + x) & \ = x^{2} \end{matrix}$$
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