Question

For a binomial distribution the mean and variance are respectively $4$ and $3$. The probability of getting a non-zero success is

• A. ${\left(\frac{1}{4}\right)}^{16}$
• B. ${\left(\frac{3}{4}\right)}^{16}$
• C. $1-{\left(\frac{3}{4}\right)}^{16}$
• D. $1-{\left(\frac{1}{4}\right)}^{16}$

Answer 1

The correct option is C $1-{\left(\frac{3}{4}\right)}^{16}$

$Mean=np=4$
$Variance=np(1-p)=3$
Hence,
$4(1-p)=3$
$(1-p)=\frac{3}{4}$
$p=\frac{1}{4}$
Therefore $n=16$
Where $n$ is the no. of trials.
Hence probability of a non zero success.
$=1-(\frac{3}{4}{)}^{16}$