Of the three independent events $E_{1}, E_{2}$ and $E_{3},$ the probability that only $E_{1}$ occurs is α, only $E_{2}$ occurs is β and only $E_{3}$ occurs is γ. Let the probability p that none of $E_{1}, E_{2}$ or $E_{3}$ occurs satisfy the equations $(α−2β)p = αβ$ and $(β−3γ)p = 2βγ.$ All the given probabilities are assumed to lie in (0,1). Then $P(E_{1})/P(E_{3}) = ?$

Of the three independent events E₁, E₂ and E₃, the probability that only E₁ occurs is α, only E₂ occurs is β and only E₃ occurs is γ. Let the probability p that none of events E₁, E₂ or E₃ occurs satisfy the equations (α - 2β)p = αβ and (β - 3γ)p = 2βγ. All the given probabilities are assumed to lie in the interval (0, 1). Then (Probability of occurrence of E₁)/(Probability of occurrence of E₃) $E_1,\ E_2,\ E_3 \text{ independent}$ $P(\text{only } E_1) = \alpha,\quad P(\text{only } E_2) = \beta,\quad P(\text{only } E_3) = \gamma$ $p = P(\text{none}),\quad (\alpha - 2\beta)p = \alpha\beta,\quad (\beta - 3\gamma)p = 2\beta\gamma$ $\text{To find: } \dfrac{P(E_1)}{P(E_3)}$ $x = P(E_1),\ y = P(E_2),\ z = P(E_3)$ $\alpha,\ \beta,\ \gamma,\ p \text{ in terms of } x,\ y,\ z$ $\text{substitute in the two equations} \Rightarrow \dfrac{x}{z}$ $\alpha = P(\text{only } E_1) = x(1-y)(1-z)$ $\beta = (1-x)y(1-z),\qquad \gamma = (1-x)(1-y)z$ $p = (1-x)(1-y)(1-z)$ $\alpha - 2\beta = (1-z)\left[x(1-y) - 2(1-x)y\right]$ $= (1-z)\,(x + xy - 2y)$ $(\alpha-2\beta)p = (1-z)(x+xy-2y)\cdot(1-x)(1-y)(1-z)$ $\alpha\beta = xy\,(1-x)(1-y)(1-z)^2$ $\Rightarrow x + xy - 2y = xy \Rightarrow x = 2y$ $\beta - 3\gamma = (1-x)\left[y(1-z) - 3(1-y)z\right]$ $= (1-x)\,(y + 2yz - 3z)$ $(1-x)(y+2yz-3z)\cdot(1-x)(1-y)(1-z) = 2yz\,(1-x)^2(1-y)(1-z)$ $\Rightarrow y + 2yz - 3z = 2yz \Rightarrow y = 3z$ $x = 2y = 2(3z) = 6z$ $\dfrac{P(E_1)}{P(E_3)} = \dfrac{x}{z} = 6$ $\boxed{\dfrac{P(E_1)}{P(E_3)} = 6}$