Imagine I have a point source $p\_0 = (x\_0,y\_0,z\_0)$ that releases a point-like Brownian particle with a lifetime given by an exponentially distributed rate parameter $\lambda$. When the particle's lifetime is over, it vanishes. We can also write an expression for the mean square displacement of the particle as a function of time as: $<x^2> = A \times t \space \frac{cm^2}{s}$, where $A \in \mathbb R^+$ is some positive real valued constant (and "cm" stands for "centimeter" and "s" stands for "seconds", just to use standard units).


What is the probability that the particle moves some distance $\leq k$ cm from $p\_0$? 


Also, if I place a ball of radius $r$ at a position $p\_1$ a distance $||p\_0 - p\_1 || = d$ away from $p\_0$, what is the probability that (one instance of) the point-like Brownian particles released at $p\_0$ hits and absorbs at the sphere before its lifetime is over and it vanishes? What if we release the particle on the surface of a reflecting plane (that prevents translation of the particle to coordinates where $z \leq 0$)? In the former case without the reflecting plane, is it possible to write down an exact analytic expression for the absorption probability, or is this generally too difficult to do for these sort of questions? 


With regards to hitting time estimations, I've never actually seen a treatment in the literature for particles with finite lifetimes diffusing from a defined point source to a defined target in an infinite volume. This is actually very surprising given that this sort of situation should arise frequently in biological settings e.g. in the context of diffusion away from the surfaces of organelles and the establishment of morphogen gradients to properly regulate tissue. This is just to say that I'm surprised, and I'll stop there because this is a mathematics site. There are, nice approximations for situations where particles with infinite lifetimes initiate randomly (or hit a randomly positioned target) in a spherical finite volume. See, for example, equations (50) and (51), for three- and two-dimensional Brownian motions, respectively, on pg. 21 (and Appendix A) of Condamin et. al. (<http://arxiv.org/abs/cond-mat/0610231v2>).


Update (with regards to Bjørn Kjos-Hanssen's answer):


In this answer, it is pointed out that the requested probability can be expressed in terms of parameters $f\_X$ for the hitting time (i.e. the time for the particle to diffuse into the target sphere), and an expression for the particle's lifetime probability distribution $f\_Y$ (here we should have $f\_Y = \lambda \times e^{-\lambda x}$), as:


$\mathbb P(X>Y) = \iint\_{\{(x,y): x>y\}} f\_X(x)f\_Y(y)\, dx\, dy$ 


(Again, this is from Bjørn Kjos-Hanssen!)


So I suppose what we're looking for here are good approximations for the hitting time distribution $f\_X$, in three-dimensions, as a function of maybe $||p\_0 - p\_1|| = d$ (i.e. the source-to-target distance) and the mean square displacement of the particle versus time (i.e. $<x^2> = A \times t \space \frac{cm^2}{s}$). However, other than perhaps expecting $f\_X$ to have some dependence on $\sqrt{t}$ (regardless of the dimension of the space in which the Brownian motion is occurring), I am unsure how to proceed.



Probability a Brownian particle with an exponentially distributed lifetime hits a sphere before vanishing
**In general,** if $Y$ is an exponentially distributed "vanishing time" and $X$ is any random time independent of $Y$, then the probability of $X$ occurring before vanishing is actually the Laplace transform of the density of $X$. Namely,
$$
\mathbb P(X<Y) = \iint\_{\{(x,y): x<y\}} f\_X(x)f\_Y(y)\, dx\, dy
$$
$$
= \int\_{0}^\infty \int^{\infty}\_x f\_X(x)f\_Y(y)\, dy\, dx
$$
$$
= \int\_{0}^\infty f\_X(x)e^{-\lambda x}\, dx = (\mathcal Lf\_X)(\lambda).
$$
Thus your question reduces to finding $f\_X$ and calculating a Laplace transform.


**In particular,** if $X$ is your hitting time of Brownian motion in dimension $n=1$, then it has a [Levy distribution](http://en.wikipedia.org/wiki/L%C3%A9vy_distribution),
$$f\_X(x) =\frac{\alpha}{\sqrt{2\pi}}\frac{\exp\left[-\frac{\alpha^2}{2x}\right]}{x^{3/2}}.$$
Let us for definiteness set $\alpha=1$, so we are looking to go from the start point 0 to a sphere whose left end point is 1. 
Then the probability of hitting 1 before vanishing is exactly
$$
(\mathcal Lf\_X)(\lambda) = \exp(-\sqrt{2\lambda}),
$$
(see Wolfram Alpha or [this paper](http://arxiv.org/pdf/1102.2709.pdf)). But if $n>1$ then I don't know what $f\_X$ would be.


