1. Let the event probabilities be

$P(E_1)=x,\quad P(E_2)=y,\quad P(E_3)=z$

Since the events are independent,

• Probability that only $E_1$ occurs: $\alpha=x(1-y)(1-z)$
• Probability that only $E_2$ occurs: $\beta=(1-x)y(1-z)$
• Probability that only $E_3$ occurs: $\gamma=(1-x)(1-y)z$
• Probability that none occurs: $p=(1-x)(1-y)(1-z)$

2. Use the given relations to simplify ratios

We are given:

$(\alpha-2\beta)p=\alpha\beta$ $(\beta-3\gamma)p=2\beta\gamma$

Now express $\alpha,\beta,\gamma$ in terms of $p$.

From $p=(1-x)(1-y)(1-z),$ we get

$\alpha=x(1-y)(1-z)=\frac{x}{1-x}p$ $\beta=(1-x)y(1-z)=\frac{y}{1-y}p$ $\gamma=(1-x)(1-y)z=\frac{z}{1-z}p$

Let $a=\frac{x}{1-x},\quad b=\frac{y}{1-y},\quad c=\frac{z}{1-z}$ Then $\alpha=ap,\quad \beta=bp,\quad \gamma=cp$

3. Substitute into the first equation

$(\alpha-2\beta)p=\alpha\beta$ becomes

$(ap-2bp)p=(ap)(bp)$ $(a-2b)p^2=abp^2$ Since $p\in(0,1)$, $p^2>0$, so

$a-2b=ab$

Thus,

$a=b(a+2)$ $b=\frac{a}{a+2}$

4. Substitute into the second equation

$(\beta-3\gamma)p=2\beta\gamma$ becomes

$(bp-3cp)p=2(bp)(cp)$ $(b-3c)p^2=2bcp^2$ So,

$b-3c=2bc$

Hence,

$b=c(2b+3)$ $c=\frac{b}{2b+3}$

5. Find the required ratio

We need

$\frac{P(E_1)}{P(E_3)}=\frac{x}{z}$

Now,

$a=\frac{x}{1-x} \implies x=\frac{a}{1+a}$ $c=\frac{z}{1-z} \implies z=\frac{c}{1+c}$

So,

$\frac{x}{z}=\frac{\frac{a}{1+a}}{\frac{c}{1+c}}=\frac{a(1+c)}{c(1+a)}$

Instead of solving fully for $a,b,c$, use the relations.

From $a-2b=ab$ we get $a=\frac{2b}{1-b}$

From $b-3c=2bc$ we get $b=\frac{3c}{1-2c}$

Substitute into $a$:

$a=\frac{2\cdot \frac{3c}{1-2c}}{1-\frac{3c}{1-2c}}$

Simplify denominator:

$1-\frac{3c}{1-2c}=\frac{1-2c-3c}{1-2c}=\frac{1-5c}{1-2c}$

Therefore,

$a=\frac{\frac{6c}{1-2c}}{\frac{1-5c}{1-2c}}=\frac{6c}{1-5c}$

Now compute:

$x=\frac{a}{1+a}=\frac{\frac{6c}{1-5c}}{1+\frac{6c}{1-5c}}=\frac{6c}{1+c}$

Also,

$z=\frac{c}{1+c}$

Hence,

$\frac{x}{z}=\frac{\frac{6c}{1+c}}{\frac{c}{1+c}}=6$

6. Final answer

$\frac{P(E_1)}{P(E_3)}=6$