Let in a Binomial distribution, consisting of \( 5 \) independent trials, probabilities of exactly \( 1 \) and \( 2 \) success be \( 0.4096 \) and \( 0.2048 \) respectively. Then the probability of getting exactly \( 3 \) successes is equal to:
\( \dfrac{128}{625} \)
\( \dfrac{32}{625} \)
\( \dfrac{80}{243} \)
\( \dfrac{40}{243} \)
