The mean and variance of a binomial distribution are α and (α)/(3)...

Given mean $np = \alpha ⋯ \left(1\right)</p><p>Also given variance$npq = $\dfrac{\alpha}{3}$ ⋯ \left(2\right)$</p><p>From \left(1\right)$ and \left(2\right)$</p><p>$q = \dfrac{1}{3}$and$p = \dfrac{2}{3}$</p><p>Now$P\left(X = 1\right) = \dfrac{4}{243}$</p><p>$⇒ C1 n {q}^{n - 1}{p}^{1} = \dfrac{4}{243}$</p><p>$⇒ \dfrac{n}{{3}^{n}} = \dfrac{2}{243}$</p><p>$⇒ n = 6$</p><p>So,$P\left(4or5\right) = C4 6 {\left(\dfrac{2}{3}\right)}^{4}{\left(\dfrac{1}{3}\right)}^{2} + C5 6 {\left(\dfrac{2}{3}\right)}^{5} \cdot {\left(\dfrac{1}{3}\right)}^{0}$</p><p>$= \dfrac{16}{27}