Question

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) \(\frac{128}{625}\)

Answer 1

Correct option is (1) \(\frac{32}{625}\)