Page 71 Question 5 from Chapter 2.2 Continuous Density Functions

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 λ = 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

0.6321206

lambda <- 1
integrate(function(t) {(lambda * exp(-lambda * t))}, lower = 0, upper = 1)
## 0.6321206 with absolute error < 7e-15

(b) within the next 3 seconds.

0.9502129

lambda <- 1
integrate(function(t) {(lambda * exp(-lambda * t))}, lower = 0, upper = 3)
## 0.9502129 with absolute error < 1.1e-14

(c) between 3 and 4 seconds from now.

0.0314743

lambda <- 1
integrate(function(t) {(lambda * exp(-lambda * t))}, lower = 3, upper = 4)
## 0.03147143 with absolute error < 3.5e-16

(d) after 4 seconds from now.

0.01831564

lambda <- 1
zeroToFour <- integrate(function(t) {(lambda * exp(-lambda * t))}, lower = 0, upper = 4)
1 - zeroToFour$value
## [1] 0.01831564