1. Write the given probabilities using independence

Let $P(B_1)=x,\quad P(B_2)=y,\quad P(B_3)=z.$ Since the events are independent,

• Probability that only $B_1$ occurs: $\alpha = x(1-y)(1-z).$
• Probability that only $B_2$ occurs: $\beta = (1-x)y(1-z).$
• Probability that only $B_3$ occurs: $\gamma = (1-x)(1-y)z.$
• Probability that none occurs: $p=(1-x)(1-y)(1-z).$

2. Relate $\alpha,\beta,\gamma$ with $p$

Divide each by $p$: $\frac{\alpha}{p}=\frac{x}{1-x},\qquad \frac{\beta}{p}=\frac{y}{1-y},\qquad \frac{\gamma}{p}=\frac{z}{1-z}.$ Let $a=\frac{\alpha}{p},\quad b=\frac{\beta}{p},\quad c=\frac{\gamma}{p}.$ Then $\alpha=ap,\quad \beta=bp,\quad \gamma=cp.$

3. Use the first given equation

Given $(\alpha-2\beta)p=\alpha\beta.$ Substitute $\alpha=ap,\ \beta=bp$: $(ap-2bp)p=(ap)(bp).$ $(a-2b)p^2=abp^2.$ Since $p\neq 0$, a-2b=ab. \tag{1}

4. Use the second given equation

Given $(\beta-3\gamma)p=2\beta\gamma.$ Substitute $\beta=bp,\ \gamma=cp$: $(bp-3cp)p=2(bp)(cp).$ $(b-3c)p^2=2bcp^2.$ Hence, b-3c=2bc. \tag{2}

5. Now express $a,b,c$ in terms of $x,y,z$

From step 2, $a=\frac{x}{1-x},\quad b=\frac{y}{1-y},\quad c=\frac{z}{1-z}.$

But it is easier to solve (1), (2) directly in $a,b,c$.

From (1): $a-2b=ab$ $a(1-b)=2b$ a=\frac{2b}{1-b}. \tag{3}

From (2): $b-3c=2bc$ $b=c(3+2b)$ c=\frac{b}{3+2b}. \tag{4}

6. Find $\dfrac{P(B_1)}{P(B_3)}$

Since $a=\frac{x}{1-x} \implies x=\frac{a}{1+a},$ $c=\frac{z}{1-z} \implies z=\frac{c}{1+c}.$ Thus $\frac{P(B_1)}{P(B_3)}=\frac{x}{z}=\frac{\frac{a}{1+a}}{\frac{c}{1+c}}=\frac{a(1+c)}{c(1+a)}.$

Now use (3) and (4).

First, $a=\frac{2b}{1-b},\qquad c=\frac{b}{3+2b}.$ Then $1+a=1+\frac{2b}{1-b}=\frac{1+b}{1-b},$ $1+c=1+\frac{b}{3+2b}=\frac{3+3b}{3+2b}=\frac{3(1+b)}{3+2b}.$

So, $\frac{x}{z}=\frac{\frac{2b}{1-b}\cdot \frac{3(1+b)}{3+2b}}{\frac{b}{3+2b}\cdot \frac{1+b}{1-b}}.$ Cancel common factors $b,(1+b),(1-b),(3+2b)$: $\frac{x}{z}=\frac{2\cdot 3}{1}=6.$

Therefore, $\boxed{\frac{P(B_1)}{P(B_3)}=6}.$

7. Comparison with stored answer

Stored correct answer = $6$. Our derived answer is also $6$, so they agree.