Find $$$P{\left(X = 1 \right)}$$$ for binomial distribution with $$$n = 2$$$ and $$$p = 0.09$$$

Calculate the various values for the binomial distribution with $$$n = 2$$$, $$$p = 0.09 = \frac{9}{100}$$$, and $$$x = 1$$$.

Mean: $$$\mu = n p = \left(2\right)\cdot \left(\frac{9}{100}\right) = \frac{9}{50} = 0.18$$$A.

Variance: $$$\sigma^{2} = n p \left(1 - p\right) = \left(2\right)\cdot \left(\frac{9}{100}\right)\cdot \left(1 - \frac{9}{100}\right) = \frac{819}{5000} = 0.1638.$$$A

Standard deviation: $$$\sigma = \sqrt{n p \left(1 - p\right)} = \sqrt{\left(2\right)\cdot \left(\frac{9}{100}\right)\cdot \left(1 - \frac{9}{100}\right)} = \frac{3 \sqrt{182}}{100}\approx 0.404722126896961.$$$A

$$$P{\left(X = 1 \right)} = 0.1638$$$A

$$$P{\left(X \lt 1 \right)} = 0.8281$$$A

$$$P{\left(X \leq 1 \right)} = 0.9919$$$A

$$$P{\left(X \gt 1 \right)} = 0.0081$$$A

$$$P{\left(X \geq 1 \right)} = 0.1719$$$A