MathsGee Answers is Zero-Rated (You do not need data to access) on: Telkom | Dimension Data | Rain | MWEB
First time here? Checkout the FAQs!
x
MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

2 like 0 dislike
34 views
Using the axioms of probability, prove the following:

a. For any event $A, P\left(A^{c}\right)=1-P(A)$.
b. The probability of the empty set is zero, i.e., $P(\emptyset)=0$.
c. For any event $A$, $P(A) \leq 1$.
d. $P(A-B)=P(A)-P(A \cap B)$.
e. $P(A \cup B)=P(A)+P(B)-P(A \cap B)$, (inclusion-exclusion principle for $n=2$ ).
f. If $A \subset B$ then $P(A) \leq P(B)$.
in Data Science & Statistics by Gold Status (10,261 points) | 34 views

1 Answer

0 like 0 dislike
Best answer
a. This states that the probability that $A$ does not occur is $1-P(A)$. To prove it using the axioms, we can write
$$
\begin{aligned}
1 &=P(S) \\
&=P\left(A \cup A^{c}\right)
\end{aligned}
$$
$($ axiom 2$)$ (definition of complement) $=P(A)+P\left(A^{c}\right) \quad$ (since $A$ and $A^{c}$ are disjoint $)$
b. Since $\emptyset=S^{c}$, we can use part (a) to see that $P(\emptyset)=1-P(S)=0$. Note that this makes sense as by definition: an event happens if the outcome of the random experiment belongs to that event. Since the empty set does not have any element, the outcome of the experiment never belongs to the empty set.
c. From part (a), $P(A)=1-P\left(A^{c}\right)$ and since $P\left(A^{c}\right) \geq 0$ (the first axiom), we have $P(A) \leq 1$
d. We show that $P(A)=P(A \cap B)+P(A-B)$. Note that the two sets $A \cap B$ and $A-B$ are disjoint and their union is $A$ (Figure $1.17)$. Thus, by the third axiom of probability
$$
\begin{array}{rlr}
P(A) & =P((A \cap B) \cup(A-B)) & (\text { since } A=(A \cap B) \cup(A-B)) \\
& =P(A \cap B)+P(A-B) & (\text { since } A \cap B \text { and } A-B \text { are disjoint }) .
\end{array}
$$

Note that since $A-B=A \cap B^{c}$, we have shown
$$
P(A)=P(A \cap B)+P\left(A \cap B^{c}\right)
$$
Note also that the two sets $B$ and $B^{c}$ form a partition of the sample space (since they are disjoint and their union is the whole sample space). This is a simple form of law of total probability that we will discuss shortly and is a very useful rule in finding probability of some
events.
e. Note that $A$ and $B-A$ are disjoint sets and their union is $A \cup B$. Thus,
$$
\begin{aligned}
P(A \cup B) &=P(A \cup(B-A)) & &(A \cup B=A \cup(B-A)) \\
&=P(A)+P(B-A) & &(\text { since } A \text { and } B-A \text { are disjoint) }
\end{aligned}
$$
$$
=P(A)+P(B)-P(A \cap B) \quad \text { (by part }(\mathrm{d}))
$$
f. Note that $A \subset B$ means that whenever $A$ occurs $B$ occurs, too. Thus intuitively we expect that $P(A) \leq P(B)$. Again the proof is similar as before. If $A \subset B$, then $A \cap B=A$. Thus,
$$
P(B)=P(A \cap B)+P(B-A)
$$
(by part (d))
$$
\begin{array}{ll}
=P(A)+P(B-A) & (\text { since } A=A \cap B) \\
\geq P(A) & (\text { by axiom } 1)
\end{array}
$$
by Gold Status (10,261 points)

Related questions

0 like 0 dislike
1 answer
asked May 22 in Mathematics by MathsGee Diamond (74,866 points) | 13 views
0 like 0 dislike
1 answer
asked May 3 in Mathematics by MathsGee Diamond (74,866 points) | 22 views
0 like 0 dislike
1 answer
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

MathsGee provides answers to subject-specific educational questions for improved outcomes.



On MathsGee Answers, you can:


1. Ask questions
2. Answer questions
3. Comment on Answers
4. Vote on Questions and Answers
5. Donate to your favourite users

MathsGee Tools

Math Worksheet Generator

Math Algebra Solver

Trigonometry Simulations

Vectors Simulations

Matrix Arithmetic Simulations

Matrix Transformations Simulations

Quadratic Equations Simulations

Probability & Statistics Simulations

PHET Simulations

Visual Statistics

ZeroEd Search Engine

Other Tools

MathsGee ZOOM | eBook