In 5 trials of binomial distribution, the probability of 3 successes is 4 times the probability of 2 successes. The probability of success in each trial is:

In 5 trials of a binomial distribution, the probability of 3 successes is 4 times the probability of 2 successes, where $n = 5$ and $p$ is the probability of success in each trial.

For binomial distribution, the probability of exactly $k$ successes is:

$P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}$

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The given condition states that $P(3) = 4 \times P(2)$:

$\binom{5}{3} \times p^3 \times (1-p)^2 = 4 \times \binom{5}{2} \times p^2 \times (1-p)^3$

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Calculating the combinations:

$\binom{5}{3} = \dfrac{5!}{3! \times 2!} = 10$

$\binom{5}{2} = \dfrac{5!}{2! \times 3!} = 10$

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Substituting the combination values:

$10 \times p^3 \times (1-p)^2 = 4 \times 10 \times p^2 \times (1-p)^3$

$p^3 \times (1-p)^2 = 4 \times p^2 \times (1-p)^3$

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Dividing both sides by $p^2 \times (1-p)^2$:

$p = 4 \times (1-p)$

$p = 4 - 4p$

$5p = 4$

$p = \dfrac{4}{5}$

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Therefore, the probability of success in each trial is $\dfrac{4}{5}$.