Mean and variance of a binomial distribution are and 3.

Question

Mean and variance of a binomial distribution are $\alpha$ and $\dfrac{\alpha}{3}$. If $P(X=1)=\dfrac{4}{243}$, find $P(X=4	ext{ or }5)$.

Accepted Answer

$q=\dfrac13,p=\dfrac23$. From $np\cdot q^{n-1}=\dfrac{4}{243}\Rightarrow n=6$.

$$P(4\text{ or }5) = {}^6C_4\left(\frac23\right)^4\left(\frac13\right)^2+{}^6C_5\left(\frac23\right)^5\left(\frac13\right) = \frac{16}{27}$$

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