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