Question

For a $B.D.$ $\overline{x}=4,\sigma =\sqrt{3}$, then $P(X=r)$=

• A. ${}^{16}{C}_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{16-r}$
• B. ${}^{12}{C}_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{12-r}$
• C. ${}^{12}{C}_r{\left(\frac{2}{3}\right)}^r{\left(\frac{1}{3}\right)}^{12-r}$
• D. ${}^{12}{C}_r{\left(\frac{3}{4}\right)}^r{\left(\frac{1}{4}\right)}^{12-r}$

Answer 1

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

$Mean=np=4$
$Variance=(\sigma {)}^2=3$
$np(1-p)=3$
$4(1-p)=3$
$1-p=\frac{3}{4}$
$p=\frac{1}{3}$

$n=4\left(4\right)$

$=16$
Therefore $P(X=r)$ implies
${}^n{C}_r\left(\frac{1}{4}{)}^r\right(\frac{3}{4}{)}^{n-r}$

$={}^{16}{C}_r\left(\frac{1}{4}{)}^r\right(\frac{3}{4}{)}^{16-r}$
Here $n$ no. f trials, and $p$ is the probability of success.