The problem states that the probability of an event occurring at least once in three independent trials is $\frac{19}{27}$. We are asked to find the probability of the event occurring in a single trial, expressing the result as an irreducible fraction $p/q$ with integers $p$ and $q$.

1. Problem Description

The problem states that the probability of an event occurring at least once in three independent trials is $\frac{19}{27}$. We are asked to find the probability of the event occurring in a single trial, expressing the result as an irreducible fraction $p/q$ with integers $p$ and $q$.

2. Solution Steps

Let $p$ be the probability of the event occurring in a single trial.

Then, $1 - p$ is the probability of the event not occurring in a single trial.

Since the trials are independent, the probability of the event not occurring in three consecutive trials is $(1-p)^3$.

The probability of the event occurring at least once in three trials is $1 - (1-p)^3$.

We are given that this probability is $\frac{19}{27}$. Therefore,

$1 - (1-p)^3 = \frac{19}{27}$

$(1-p)^3 = 1 - \frac{19}{27} = \frac{27 - 19}{27} = \frac{8}{27}$

Taking the cube root of both sides:

$1 - p = \sqrt[3]{\frac{8}{27}} = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} = \frac{2}{3}$

$p = 1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$

3. Final Answer

The probability of the event occurring in a single trial is $\frac{1}{3}$.