Question

Construct the probability distribution function (PDF). Stop at $x=6$.

$\begin{array}{|ll|}\hline x& P\left(x\right)\\ 1& \\ 2& \\ 3& \\ 4& \\ 5& \\ 6& \\ \hline\end{array}$

Short answer

The constructed table for the PDF is

$\begin{array}{|ll|}\hline X& P\left(X\right)\\ 1& 0.713\\ 2& 0.20463\\ 3& 0.05873\\ 4& 0.01686\\ 5& 0.00484\\ 6& 0.00139\\ \hline\end{array}$

Step-by-step solution

Given Information

Success probability $p=0.713$.

Explanation

Formula :

$P(X=x)=p(1-p{)}^{x-1};0\le p\le 1,x=1,2,\dots$

When $x=1,2,3,4,5,6$is substituted into the preceding formula, the required probability distribution function (PDF) for $X$is:

$\begin{array}{|ll|}\hline X& P\left(X\right)\\ 1& 0.713\\ 2& 0.20463\\ 3& 0.05873\\ 4& 0.01686\\ 5& 0.00484\\ 6& 0.00139\\ \hline\end{array}$