If the probability that the random variable X takes the value x is given by \(\mathrm{P}(\mathrm{X}=\mathrm{x})=\mathrm{…

We have

\(P(X=x)=k(x+1){3}^{−x},\ x=0,1,2,\ldots\)

Use \({\sum }_{x=0}^{\mathrm{∞}}(x+1){r}^{x}=\frac{1}{(1−r{)}^{2}}\) (for \(\mathrm{∣}r\mathrm{∣}<1\)) with \(r=\frac{1}{3}\):

\(\sum _{x=0}^{\mathrm{∞}}(x+1){(\frac{1}{3})}^{x}=\frac{1}{{(1−\frac{1}{3})}^{2}}=\frac{1}{{(\frac{2}{3})}^{2}}=\frac{9}{4}\)

Since total probability is 1:

\(k⋅\frac{9}{4}=1\Rightarrow k=\frac{4}{9}\)

Now

\(P(X\leq 2)=\sum _{x=0}^{2}k(x+1){3}^{−x}=\frac{4}{9}[1+\frac{2}{3}+\frac{3}{9}]=\frac{4}{9}⋅2=\frac{8}{9}\)

So

\(P(X\geq 3)=1−P(X\leq 2)=1−\frac{8}{9}=\frac{1}{9}\mathrm{.}\)