Determine the binomial distribution whose mean is 9 and whose standard deviation is $\frac{3}{2}$. Find the probability of obtaining atmost one success.

$37(\frac{1}{4})^{12}$

The correct answer is Option (2) → $37(\frac{1}{4})^{12}$

Let the binomial distribution be $(q+p)^n$.

Given that mean = $np = 9$ and standard deviation = $\sqrt{npq}=\frac{3}{2}$

$⇒\sqrt{9q} = \frac{3}{2} ⇒3\sqrt{9} = \frac{3}{2}⇒q = \frac{1}{4}$

$∴p=1-q=1-\frac{1}{4}=\frac{3}{4}$

and $np=9⇒n×\frac{3}{4}=9⇒n=12$

Hence, the binomial distribution is $(q + p)^n$ i.e. $\left(\frac{1}{4}+\frac{3}{4}\right)^{12}$

P(atmost one success) = $P(X ≤ 1) = P(0) + P(1)$

$={^{12}C}_0p^0q^{12}+{^{12}C}_1p^1q^{11}$

$=1.\left(\frac{1}{4}\right)^{12}+12.\frac{3}{4}\left(\frac{1}{4}\right)^{11}=37(\frac{1}{4})^{12}$