Find $$$P{\left(X = 3 \right)}$$$ for geometric distribution with $$$n = 3$$$ and $$$p = 0.2$$$

Calculate the various values for the geometric distribution with $$$n = 3$$$ and $$$p = 0.2 = \frac{1}{5}$$$ (don't include a success trial).

Mean: $$$\mu = \frac{1 - p}{p} = \frac{1 - \frac{1}{5}}{\frac{1}{5}} = 4$$$A.

Variance: $$$\sigma^{2} = \frac{1 - p}{p^{2}} = \frac{1 - \frac{1}{5}}{\left(\frac{1}{5}\right)^{2}} = 20$$$A.

Standard deviation: $$$\sigma = \sqrt{\frac{1 - p}{p^{2}}} = \sqrt{\frac{1 - \frac{1}{5}}{\left(\frac{1}{5}\right)^{2}}} = 2 \sqrt{5}\approx 4.472135954999579.$$$A

$$$P{\left(X = 3 \right)} = 0.1024$$$A

$$$P{\left(X \lt 3 \right)} = 0.488$$$A

$$$P{\left(X \leq 3 \right)} = 0.5904$$$A

$$$P{\left(X \gt 3 \right)} = 0.4096$$$A

$$$P{\left(X \geq 3 \right)} = 0.512$$$A