In the previous section, we saw Exhaustive events. In this section, we will see Axiomatic approach to probability.
• In our earlier classes, we have seen how probability can be calculated in simple cases.
• We calculated probability as a fraction. It was a proper fraction. That is., numerator will be less than the denominator. If a proper fraction is converted to decimal form, the value on the left side of the decimal point will be zero.
• In the denominator, we gave total number of possible outcomes. In the numerator, we gave the total number of favorable outcomes.
• In our present discussion, we will see three axioms developed by the Russian mathematician A.N. Kolmogorov. Those axioms will enable us to calculate probability in complex problems.
• Axiom means a statement or rule, which is self-evident. Such a
statement/rule can be a useful basis for more advanced mathematical
calculations.
Axiom 1: For any event A, P(A) ≥ 0.
• This can be explained in 5 steps:
(i) The probability any event A to occur will be greater than or equal to zero.
♦ If P(A) is zero, A will never occur.
♦ If P(A) is large, there is a higher probability for A to occur.
♦ If P(A) is small, there is a lower probability for A to occur.
(ii) We saw that, probability is always a proper fraction. So the largest possible value for P(A) is 1. When the probability is 1, the event will surely occur.
(iii) Also, the smallest possible value for P(A) is zero. When the probability is zero, the event will never occur.
(iv) Since P(A) is a proper fraction, “zero value” is obtained when the numerator is zero. This happens when there are no favorable outcomes.
(v) Since P(A) is a proper fraction, “1 value” is obtained when the numerator is same as the denominator. This happens when number of favorable outcomes is same as the total number of outcomes.
Axiom 2:
For any experiment, P(S) = 1
• This can be explained in 2 steps:
(i) We know that, any event A can be denoted as a set A. The elements of set A will be the favorable outcomes of event A.
(ii) The sample space of an experiment is also a set. We denote it as S.
• So this S can be considered as an event. Whenever we perform that experiment, we will surely obtain an element of S. Based on this, we can say: Probability for S to occur is 1.
Axiom 3:
If A and B are two mutually exclusive events then P(A∪B) = P(A) + P(B)
• This can be explained using an example. It can be written in 5 steps:
(i) First we will assign the sets.
• Let set S of an experiment have 6 elements.
• Let event A of that experiment have 3 elements.
• Let event B of that experiment have 2 elements.
(ii) Now we will write the probabilities of the individual events:
• We get:
♦ P(A) = 3/6 = 1/2
♦ P(B) = 2/6 = 1/3
(iii) Given that A and B are disjoint sets. So A∪B will have 5 elements.
• So P(A∪B) = 5/6
(iv) Let us add the individual probabilities:
P(A) + P(B) = 1/2 + 1/3 = 5/6
• This is same as the result in (iii)
(v) Comparing the results in (iii) and (iv), we can write:
P(A∪B) = P(A) + P(B)
Using the above axioms, we can derive four important results.
Result 1:
This can be written in 3 steps:
1. Suppose that, B is a null set. That means, no outcome in S can be used to define the event B.
(For example, when a die is thrown once, the event of “getting number 7” is a null set)
• So we can write: B = ะค
2. Applying axiom 3, we get:
$\begin{array}{ll}
{}&{P(A \cup B)}
& {~=~}& {P(A) + P(B)}
&{} \\
{\Rightarrow}&{P(A \cup \phi)}
& {~=~}& {P(A) + P(\phi)~~ \color {green} {\text{- - - (I)}}}
&{} \\
{\Rightarrow}&{P(A)}
& {~=~}& {P(A) + P(\phi)~~ \color {green} {\text{- - - (II)}}}
&{} \\
{\Rightarrow}&{P(\phi)}
& {~=~}& {0}
&{} \\
\end{array}$
◼ Remarks:
• Line marked as I:
In this line, we are able to apply axiom 3 because, any set A and the null set are disjoint sets
• Line marked as II:
In this line, we use the rule:
Union of any set A with the null set will give A
3. Based on the above calculations, we can write:
If there are no outcomes for an event, then the probability for that event to occur is zero.
Result 2:
This can be written in 4 steps:
1. Let S be the sample space of an experiment.
• Also, let ๐1, ๐2, ๐3, . . . , ๐n be the n elements of S.
• Then we can write: S= {๐1, ๐2, ๐3, . . . , ๐n}
2. So S is the union of n singleton sets:
S = {๐1} ∪ {๐2} ∪ {๐3} ∪ . . . ∪ {๐n}
3. Applying axiom 3, we get:
$\begin{array}{ll}
{}&{\rm{P \left(\lbrace \omega_1 \rbrace
\cup \lbrace \omega_2 \rbrace \cup \lbrace \omega_3 \rbrace \cup ~.~.~.~
\cup \lbrace \omega_n \rbrace \right)}}
& {~=~}& {\rm{P
\left(\lbrace \omega_1 \rbrace \right) + P \left(\lbrace \omega_2
\rbrace \right) + P \left(\lbrace \omega_3 \rbrace \right) ~+~.~.~.~+~P
\left(\lbrace \omega_n \rbrace \right)}}
&{} \\
{\Rightarrow}&{\rm{P(S)}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P
\left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3
\rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\
{\Rightarrow}&{\rm{1}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P
\left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3
\rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\
\end{array}$
◼ Remarks:
From axiom 2, we know that: P(S) = 1
4. So we can write:
Sum of the individual probabilities of all individual outcomes of an experiment will be 1.
Result 3:
This can be written in 5 steps:
1. From result 2, we know that:
P({๐1}) + P({๐2}) + P({๐3}) + . . . + P({๐n}) = 1
• Note that, the sum on the RHS is 1.
2. Each term on the LHS is a proper fraction.
• Since the sum is 1, each term must be either less than 1 or equal to 1.
3. But axiom 1 tells us that:
Each term will be either greater than zero or equal to zero.
4. Combining (2) and (3), we get:
0 ≤ Any term ≤ 1
• Mathematically, we write this as:
0 ≤ P({๐i}) ≤ 1
5. So we can write:
Take any outcome from S. That outcome can be considered as a simple event (event with only one outcome). The probability of that event will be greater than or equal to zero and at the same time, less than or equal to 1.
Result 4:
This can be written in 4 steps:
1. Let A be an event in an experiment.
• Also, let ๐1, ๐2, ๐3, . . . , ๐n be the n elements of A.
• Then we can write: A = {๐1, ๐2, ๐3, . . . , ๐n}
2. So A is the union of n singleton sets:
A = {๐1} ∪ {๐2} ∪ {๐3} ∪ . . . ∪ {๐n}
3. Applying axiom 3, we get:
$\begin{array}{ll}
{}&{\rm{P \left(\lbrace \omega_1 \rbrace
\cup \lbrace \omega_2 \rbrace \cup \lbrace \omega_3 \rbrace \cup ~.~.~.~
\cup \lbrace \omega_n \rbrace \right)}}
& {~=~}& {\rm{P
\left(\lbrace \omega_1 \rbrace \right) + P \left(\lbrace \omega_2
\rbrace \right) + P \left(\lbrace \omega_3 \rbrace \right) ~+~.~.~.~+~P
\left(\lbrace \omega_n \rbrace \right)}}
&{} \\
{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{P \left(\lbrace \omega_1 \rbrace \right) + P
\left(\lbrace \omega_2 \rbrace \right) + P \left(\lbrace \omega_3
\rbrace \right) ~+~.~.~.~+~P \left(\lbrace \omega_n \rbrace \right)}}
&{} \\
{\Rightarrow}&{\rm{P(A)}}
& {~=~}& {\rm{\sum{P \left(\lbrace \omega_i \rbrace \right)}}}
&{} \\
\end{array}$
4. So we can write:
Sum of the individual probabilities of all individual outcomes of an event will be the probability of that event.
◼ Based on the three axioms and four results, we can write:
Any experiment must satisfy two conditions:
Condition 1:
Take any individual outcome of an experiment. The probability of that outcome will be greater than or equal to zero, but at the same time less than or equal to 1.
•
Mathematically, we can write this as:
$\rm{0 \le P(\lbrace \omega_i\rbrace) \le 1,~Where~ \omega_i \in S}$
Condition 2:
Take the sum of the probabilities of all individual outcomes in S. That sum will be 1.
•
Mathematically, we can write this as:
$\rm{\sum{P(\lbrace \omega_i\rbrace)} = 1,~Where~ \omega_i \in S}$
• Let us see an example. It can be written in steps:
1. Consider the experiment of tossing a coin once.
We know that, S = {H,T}
2. Let us assign individual probabilities for the two outcomes:
♦ P({H}) = ½
♦ P({T}) = ½
3. Does the above probability values satisfy the two conditions?
(i) We have: 0 ≤ ½ ≤ 1
So condition 1 is satisfied.
(ii) We have: (½ + ½) = 1
So condition 2 is also satisfied.
4. We know that exact ½ cannot be obtained even if we repeat the experiment 50000 times. What if we repeat the experiment only 20 times and get the following values?
♦ P({H}) = 1/4
♦ P({T}) = 3/4
5. Does the above probability values satisfy the two conditions?
(i) We have:
♦ 0 ≤ 1/4 ≤ 1
♦ 0 ≤ 3/4 ≤ 1
So condition 1 is satisfied.
(ii) We have: (1/4 + 3/4) = 1
So condition 2 is also satisfied.
6. We see that ¼ and ¾ also satisfy the two conditions.
• In fact, there are infinite number of possible values such that:
♦ P({H}) = p
♦ P({T}) = (1- p)
Where 0 ≤ p ≤ 1
Now we will see a solved example:
Solved example 16.9
Let a sample space be S = {๐1, ๐2, ๐3, . . . , ๐6}. Which of the following assignments of probabilities to each outcome are valid ?
$\begin{array}{cc}
{\textbf{Outcomes}}&{\mathbf{\omega_1}}
& {\mathbf{\omega_2}}& {\mathbf{\omega_3}}
&{\mathbf{\omega_4}} &{\mathbf{\omega_5}} &{\mathbf{\omega_6}}\\
{\textbf{(a)}}&{\frac{1}{6}}
& {\frac{1}{6}}& {\frac{1}{6}}
&{\frac{1}{6}} &{\frac{1}{6}} &{\frac{1}{6}}\\
{\textbf{(b)}}&{1}
& {0}& {0}
&{0} &{0} &{0}\\
{\textbf{(c)}}&{\frac{1}{8}}
& {\frac{2}{3}}& {\frac{1}{3}}
&{\frac{1}{3}} &{- \frac{1}{4}} &{- \frac{1}{3}}\\
{\textbf{(d)}}&{\frac{1}{12}}
& {\frac{1}{12}}& {\frac{1}{6}}
&{\frac{1}{6}} &{\frac{1}{6}} &{\frac{3}{2}}\\
{\textbf{(e)}}&{0.1}
& {0.2}& {0.3}
&{0.4} &{0.5} &{0.6}\\
\end{array}$
Solution:
We have two conditions to check whether the probabilities assigned to the outcomes are valid.
Condition 1:
$\rm{0 \le P(\lbrace \omega_i\rbrace) \le 1,~Where~ \omega_i \in S}$
Condition 2:
$\rm{\sum{P(\lbrace \omega_i\rbrace)} = 1,~Where~ \omega_i \in S}$
Let us check each case:
Part (a):
•
Condition 1 is satisfied because, all probability values are greater than or equal to 0 and at the same time, less than or equal to 1.
•
Condition 2 is satisfied because:
$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}~=~1$
•
Both conditions are satisfied. So the assigned probabilities are valid.
Part (b):
•
Condition 1 is satisfied because, all probability values are greater than or equal to 0 and at the same time, less than or equal to 1.
•
Condition 2 is satisfied because:
1 + 0 + 0 + 0 + 0 + 0 = 1
•
Both conditions are satisfied. So the assigned probabilities are valid.
Part (c):
•
Condition 1 is not satisfied because, P({๐5}) and P({๐6}) are less than 0.
•
Since condition 1 is not satisfied, there is no need to check condition 2.
•
Condition 1 is not satisfied. So the assigned probabilities are not valid.
Part (d):
•
Condition 1 is not satisfied because, P({๐6}) is greater than 1.
•
Since condition 1 is not satisfied, there is no need to check condition 2.
•
Condition 1 is not satisfied. So the assigned probabilities are not valid.
Part (e):
•
Condition 1 is satisfied because, all probability values are greater
than or equal to 0 and at the same time, less than or equal to 1.
•
Condition 2 is not satisfied because:
0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1 ≠ 1
•
Condition 2 is not satisfied. So the assigned probabilities are not valid.
In the next section, we will see Probability of an event.
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