Question

The mean and variance of a binomial distribution are α and α/3, respectively.
If P(X = 1) = 4/243 then P(X = 4 or 5) is equal to :

• A. $\frac{5}{9}$
• B. $\frac{64}{81}$
• C. $\frac{16}{27}$
• D. $\frac{145}{243}$

Answer 1

Answer: (C)

$\frac{16}{27}$

Answer 2

Given, mean = np = α.
variance
$=npq =\frac{α}{3}$
$⇒q=\frac{1}{3} and\ p=\frac{2}{3}$
$P(X=1)=n.p^1.q^{n−1}=\frac{4}{243}$
$⇒n.\frac{2}{3}.(\frac{1}{3})^{n−1}=\frac{4}{243}$
$⇒n=6$
$P(X=4 or 5)=^6C_4.(\frac{2}{3})^4⋅(\frac{1}{3})^2+^6C_5.(\frac{2}{5})^5.\frac{1}{3}$
$=\frac{16}{27}$
So, the correct option is (C): $\frac{16}{27}$