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Section 1.3

Know how to compute the probability with infinitely many outcomes

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Section 1.3 Infinitely many outcomes

$LaTeX: \Omega=\{1,2,3, \cdots\}$

$LaTeX: P(A)=\frac{\# A}{\# \Omega}$

Geometric Probability Distribution

Sample Space : disk of radius 9

$LaTeX: \Omega=\left\{(x, y): x^{2}+y^{2} \leq 9^{2}\right\} \text {. }$

Event LaTeX: A : disk of radius $LaTeX: \frac{1}{4}$

$LaTeX: A=\left\{(x, y): x^{2}+y^{2} \leq(1 / 4)^{2}\right\}$

$LaTeX: P(A)=\frac{\text { area of } A}{\text { area of } \Omega}=\frac{\pi \cdot(1 / 4)^{2}}{\pi \cdot 9^{2}}=\frac{1}{36^{2}} \approx 0.00077$

We pick a real number uniformly at random from the closed unit interval [0,1]. Let X denote the number chosen. "Uniformly at random" means that X is equally likely to lie anywhere in [0,1]. Obviously $LaTeX: \Omega$ = [0,1]. What is the probability that X lies in a smaller interval [a,b] $LaTeX: \subseteq$ [0,1]?

$LaTeX: \Omega=\{1,2,3, \cdots\}$

$LaTeX: \text { Event } \mathrm{A}=\{k\} \Rightarrow P(A)=?$

$LaTeX: \Omega=\{1,2,3, \cdots\}$

$LaTeX: \text { Event } \mathrm{A}=\{k\} \Rightarrow P(A)=0$

Pf) Let LaTeX: P(A)=c>0

Then $LaTeX: P(\Omega)=\sum_{k=1}^{\infty} P(k)=\sum_{k=1}^{\infty} c=c \cdot \infty=\infty \neq 1$

This is a contradiction.

Hence LaTeX: P(A)=0

Notes