Question

If the mean and variance of a binomial variate $X$ are respectively $\frac{35}{6}$and $\frac{35}{36}$, then the probability of $X>6$ is

• A. $1-{\left(\frac{5}{6}\right)}^7$
• B. ${\left(\frac{5}{6}\right)}^7$
• C. $1-{\left(\frac{1}{6}\right)}^7$
• D. ${\left(\frac{1}{6}\right)}^7$

Answer 1

Answer: (B)

${\left(\frac{5}{6}\right)}^7$

Here, $np=\frac{35}{6},npq=\frac{35}{36}$
$\therefore q=\frac{1}{6},p=\frac{5}{6},n=7$
Therefore $P(X>6)=P(X=7){=}^7{C}_7{p}^7{q}^{7-7}={\left(\frac{5}{6}\right)}^7$