Discussion Week 4

Introduction to Probability Chapter 2 #5

Suppose you are watching a radioactive source that emits particles at a rate described by the exponential density \[f(t)=\lambda e^{-\lambda t}\] where \(\lambda= 1\), so that the probability \(P(0, T )\) that a particle will appear in the next \(T\) seconds is \(P ([0, T ]) = \int _{0}^{T} \lambda e ^{−\lambda t} dt\). Find the probability that a particle (not necessarily the first) will appear:
(a) within the next second
\[P([0,1]) = \int _{0}^{1} \lambda e ^{−\lambda t} dt = \int _{0}^{1} 1 e ^{−1 t} dt = -e^{-t|_{0}^{1}} = -e^{-1}-(-e^{0}) = 1-\frac{1}{e}\]
(b) within the next 3 seconds \[P([0,3]) = \int _{0}^{3} \lambda e ^{−\lambda t} dt = \int _{0}^{3} 1 e ^{−1 t} dt = -e^{-t|_{0}^{3}} = -e^{-3}-(-e^{0}) = 1-\frac{1}{e^3}\]
(c) between 3 and 4 seconds from now \[P([3,4]) = \int _{3}^{4} \lambda e ^{−\lambda t} dt = \int _{3}^{4} 1 e ^{−1 t} dt = -e^{-t|_{3}^{4}} = -e^{-4}-(-e^{-3}) = \frac{1}{e^3} - \frac{1}{e^4}\]
(d) after 4 seconds from now \[P([4,\infty ]) = 1 - \int _{0}^{4} \lambda e ^{−\lambda t} dt = 1 - \int _{0}^{4} 1 e ^{−1 t} dt = 1 -e^{-t|_{0}^{4}} \] \[= 1 - ( -e^{-4}-(-e^{0}) ) = 1 - ( 1-\frac{1}{e^4} ) = \frac{1}{e^4}\]