Home page for accesible maths 3 The axiomatic approach 3.1 The axioms of probability 3.3 Conditional probability

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3.2 The laws of probability

Those statements of probability known as rules, lemmas, theorems or laws, are statements that can be derived by mathematical deduction from the axioms. They are consequences of the axioms and while the axioms themselves are accepted on faith, the laws need to be justified.

Theorem 3.4 (The law of complementary events).

{{\rm P}(A^{c})=1-{\rm P}(A)}.

Proof.

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${\Omega=A\cup A^{c}}$ [either the event ${A}$ or the event ${A^{c}}$ must happen, they are exhaustive];
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Using axiom 2 we have ${1={\rm P}(\Omega)={\rm P}(A\cup A^{c})}$ .
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But ${{\rm P}(A\cup A^{c})={\rm P}(A)+{\rm P}(A^{c})}$ by axiom 3, as ${A}$ and ${A^{c}}$ are exclusive.
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So ${1={\rm P}(A)+{\rm P}(A^{c})}$ .
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Rearranging this expression completes the proof.

∎

Exercise 3.5.

A fair coin is thrown ${3}$ times. What is the probability that at least one ${H}$ occurs?

Solution.

Let ${A}$ be the event that at least one H occurs.
Then ${A^{c}}$ is the event no Hs occur.
${{\rm P}(A^{c})={\rm P}(\mbox{no }H)={\rm P}(\{TTT\})=1/8}$ .
${{\rm P}(A)=1-{\rm P}(A^{c})=1-1/8=7/8}$ .

Exercise 3.6.

Show that ${{\rm P}(\emptyset)=0}$ .

Solution.

By the law of complementary events and axiom 2,

$\displaystyle{\rm P}(\emptyset)$	$\displaystyle=$	$\displaystyle 1-{\rm P}(\emptyset^{c})$
	$\displaystyle=$	$\displaystyle 1-{\rm P}(\Omega)$
	$\displaystyle=$	$\displaystyle 1-1=0.$

The law of complementary events is a useful way to consider single properties. However it provides no mechanism for dealing with two events simultaneously, such as an individual being numerate and literate. The first useful law for combining knowledge about more than one event is:

Theorem 3.7 (The partition law).

{{\rm P}(A)={\rm P}(A\cap B)+{\rm P}(A\cap B^{c})}.

Proof.

Consider the Venn diagram:

Write ${A=(A\cap B)\cup(A\cap B^{c})}$ .
Note also that $(A\cap B)\cap(A\cap B^{c})=\emptyset$ .
So ${A\cap B}$ and ${A\cap B^{c}}$ are exclusive.
Using axiom 3,

{{\rm P}(A)={\rm P}(A\cap B)+{\rm P}(A\cap B^{c}).}

∎

Exercise 3.8.

If a randomly selected individual is

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illiterate and innumerate with probability ${0.1}$ ,
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illiterate and numerate with probability ${0.2}$ ,
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literate and innumerate with probability ${0.3}$ and is
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literate and numerate with probability ${0.4}$ ,

then find the probability the individual selected is literate, and the probability the individual is illiterate.

Solution.

Let ${A}$ denote the event that the selected individual is literate and ${B}$ denote numerate.

$\displaystyle{\rm P}(\mbox{literate})$	$\displaystyle=$	$\displaystyle{\rm P}(A)$
	$\displaystyle=$	$\displaystyle{\rm P}(A\cap B)+{\rm P}(A\cap B^{c})$
	$\displaystyle=$	$\displaystyle 0.4+0.3=0.7$
$\displaystyle{\rm P}(\mbox{illiterate})$	$\displaystyle=$	$\displaystyle{\rm P}(A^{c})$
	$\displaystyle=$	$\displaystyle 1-{\rm P}(A)=0.3.$

Of all the laws of probability the addition law is the best known:

Theorem 3.9 (The addition law).

{{\rm P}(A\cup B)={\rm P}(A)+{\rm P}(B)-{\rm P}(A\cap B)}.

Proof.

Consider the Venn diagram:

We can see that

{A\cup B=A\cup(B\cap A^{c}).}

As ${A}$ and ${B\cap A^{c}}$ are exclusive, using axiom 3,

\displaystyle{\rm P}(A\cup B)={\rm P}(A)+{\rm P}(B\cap A^{c}).

But by the partition law

\displaystyle{\rm P}(B)={\rm P}(B\cap A)+{\rm P}(B\cap A^{c}).

So substituting

\displaystyle{\rm P}(A\cup B)={\rm P}(A)+{\rm P}(B)-{\rm P}(B\cap A).

∎

Exercise 3.10.

A fair die is thrown twice. Let the event ${A}$ denote an even number on the first throw, and let ${B}$ denote an even number on the second. Find the probability of at least one even number.

A diagrammatic representation of the sample space makes this clearer:

second throw:		2	4	6	1	3	5
first throw	5	b	b	b	*	*	*
	3	b	b	b	*	*	*
	1	b	b	b	*	*	*
	6	ab	ab	ab	a	a	a
	4	ab	ab	ab	a	a	a
	2	ab	ab	ab	a	a	a

Each point is equally probable.

Solution.

$\displaystyle{\rm P}(A\cup B)$	$\displaystyle=$	$\displaystyle{\rm P}(A)+{\rm P}(B)-{\rm P}(B\cap A)$
	$\displaystyle=$	$\displaystyle 18/36+18/36-9/36$
	$\displaystyle=$	$\displaystyle 3/4.$

The laws of probability are well illustrated in a ${2\times 2}$ table:

\begin{array}[]{ r|cc|c|l }&B&B^{c}&\mbox{total}&\mbox{law}\\ \hline A&{\rm P}(A\cap B)&{\rm P}(A\cap B^{c})&{\rm P}(A)&\mbox{partition}\\ A^{c}&{\rm P}(A^{c}\cap B)&{\rm P}(A^{c}\cap B^{c})&{\rm P}(A^{c})&\\ \hline\mbox{total}&{\rm P}(B)&{\rm P}(B^{c})&1&\mbox{complementary events}\end% {array}

Exercise 3.11.

Show that ${{\rm P}(A^{c}\cap B^{c})=1-{\rm P}(A)-{\rm P}(B)+{\rm P}(A\cap B)}$ .

[Hint: $A^{c}\cap B^{c}=(A\cup B)^{c}$ ]

Solution.

$\displaystyle{\rm P}(A^{c}\cap B^{c})$	$\displaystyle=$	$\displaystyle{\rm P}((A\cup B)^{c})$
	$\displaystyle=$	$\displaystyle 1-{\rm P}(A\cup B),$
	$\displaystyle=$	$\displaystyle 1-{\rm P}(A)-{\rm P}(B)+{\rm P}(B\cap A).$