I am wondering if I did this right. Compute the following
Computing
$prob(x\ge 4|x\ge 1)$
using the geometric distribution
The geometric distribution is $f(x)=p(1-p)^x$
So then would I do the following
$prob(x\ge 4|x\ge 1)=prob(x \ge 4-1)$
by the memory less property so then find $prob(x \ge 3)$
and I get
$\int_3^{\infty}p(1-p)^x$
$=\frac{p(1-p)^x}{\log(1-p)}$ from 3 to infinity and then I get
$=\frac{(p-1)^3}{\log(1-p)}$
In general it is a good idea to use the converse probability to calculate $P(X\geq x)$ if $x$ is not large-as suggested by Henry. It works perfectly here. But in case of the geometric distribution the expression for $P(X\geq x)$ is quite simple.
$$P(X\geq x)=\sum_{k=x}^{\infty} p\cdot q^{k}=p\cdot \sum_{k=x}^{\infty}q^{k}$$
$$=p\cdot \left[ \sum_{k=0}^{\infty}q^{k}-\sum_{k=0}^{x-1}q^{k} \right]= p\cdot \left[ \frac1{1-q}-\frac{1-q^x}{1-q} \right]$$
Since $p=1-q$ the denominator disappears by cancelling. Therefore
$$P(X\geq x)=q^x=(1-p)^x$$
For $p=0.3$ and $x=3$ we can check both ways:
$\texttt{a) closed forumula}$
$P(X\geq 3)=(1-0.3)^3=0.7^3$
$\texttt{b) converse probability}$
b) $P(X\geq 3)=1-P(X=2)-P(X=1)-P(X=0)=1-0.3\cdot 0.7^2-0.3\cdot 0.7-0.3$
$$=0.343=0.7^3$$