In the previous section, we saw the basics of the axiomatic approach to probability. In this section, we will see Probability of an event.
Probability of an event
This can be explained using an example. It can be written in 7 steps:
1. Consider the following experiment:
A machine continuously produces pens. Three consecutive pens are examined. Each of those three pens are classified as defective or non-defective.
2. Let us use short forms for defective and non-defective:
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We will put ‘B’ if the pen taken is defective. Here ‘B’ stands for ‘bad’.
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We will put ‘G’ if the pen taken is non-defective. Here ‘G’ stands for ‘good’.
3. So, while examining the three pens, one outcome may be: BBG
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In that case,
♦ First pen is bad.
♦ Second pen is bad.
♦ Third pen is good.
4. In this way, we can write all the possible outcomes:
S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}
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We see that, there are eight possible outcomes.
5. Next step is to assign probabilities for each of these outcomes.
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It is not mentioned that, the machine has any special properties.
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If it has the property of making mostly good pens, then the outcomes like GGG, GBG etc., will get greater probability.
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If it has the property of making mostly bad pens, then the outcomes like BBB, BGB, etc., will get greater probability.
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Since there is no mention of special properties, we will assign equal probabilities as shown below:
$\begin{array}{cc}
{\textbf{Outcomes}}&{\textbf{BBB}}
&
{\textbf{BBG}}& {\textbf{BGB}} & {\textbf{BGG}} &
{\textbf{GBB}} & {\textbf{GBG}} & {\textbf{GGB}} &
{\textbf{GGG}}\\
{\textbf{Probabilities}}&{\frac{1}{8}}
&{\frac{1}{8}} &{\frac{1}{8}} &{\frac{1}{8}}
&{\frac{1}{8}} &{\frac{1}{8}} &{\frac{1}{8}}
&{\frac{1}{8}}\\
\end{array}$
•
Note that, we put 8 in the denominator because, the total number of outcomes is 8.
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The reader may verify that, the above assigned probabilities satisfy both condition 1 and condition 2. that we saw in the previous section.
(The reader must keep in mind that, 1/8 is only a theoretical value. In actual practice, we will get “values close to 1/8” only when the experiment is repeated a very large number of times)
6. Suppose that an event A is: “getting exactly one defective pen”.
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Then we get: A = {BGG, GBG, GGB}
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Now we use result 4 that we saw in the previous section:
$\rm{P(A)~=~\sum{P \left(\lbrace \omega_i \rbrace \right)}}$
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So in our present case, we get:
$\begin{array}{ll}
{}&{\rm{P(A)}}
& {~=~}& {\rm{P \left(\lbrace BGG \rbrace \right) + P \left(\lbrace GBG \rbrace \right) + P \left(\lbrace GGB \rbrace \right)}} &{} \\
{}&{}
& {~=~}& {\frac{1}{8}~+~\frac{1}{8}~+~\frac{1}{8}} &{} \\
{}&{}
& {~=~}& {\frac{3}{8}} &{} \\
\end{array}$
7. Suppose that an event B is: “getting atleast two defective pen”.
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Then we get: B = {BBB, BBG, BGB, GBB}
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Now we use result 4 that we saw in the previous section:
$\rm{P(A)~=~\sum{P \left(\lbrace \omega_i \rbrace \right)}}$
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So in our present case, we get:
$\begin{array}{ll}
{}&{\rm{P(B)}}
&
{~=~}& {\rm{P \left(\lbrace BBB \rbrace \right) + P \left(\lbrace
BBG \rbrace \right) + P \left(\lbrace BGB \rbrace \right) + P \left(\lbrace GBB \rbrace \right)}} &{} \\
{}&{}
& {~=~}& {\frac{1}{8}~+~\frac{1}{8}~+~\frac{1}{8}~+~\frac{1}{8}} &{} \\
{}&{}
& {~=~}& {\frac{4}{8}} &{} \\
{}&{}
& {~=~}& {\frac{1}{2}} &{} \\
\end{array}$
Let us see another example. It can be written in 5 steps:
1. Consider the experiment of tossing a coin twice.
2. We know that, S = {HH, HT, TH, TT}
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We see that, there are four possible outcomes.
3. Next step is to assign probabilities for each of these outcomes.
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Let the probabilities be as shown below:
$\begin{array}{cc}
{\textbf{Outcomes}}&{\textbf{HH}}
&
{\textbf{HT}}& {\textbf{TH}} & {\textbf{TT}}\\
{\textbf{Probabilities}}&{\frac{1}{4}}
&{\frac{1}{7}} &{\frac{2}{7}} &{\frac{9}{28}}\\
\end{array}$
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The reader may verify that, the above assigned probabilities satisfy
both condition 1 and condition 2. that we saw in the previous section.
4. Suppose that an event E is: “getting the same result in both the tosses”.
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Then we get: A = {TT, HH}
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Now we use result 4 that we saw in the previous section:
$\rm{P(A)~=~\sum{P \left(\lbrace \omega_i \rbrace \right)}}$
•
So in our present case, we get:
$\begin{array}{ll}
{}&{\rm{P(E)}}
&
{~=~}& {\rm{P \left(\lbrace HH \rbrace \right) + P \left(\lbrace
TT \rbrace \right)}} &{} \\
{}&{}
& {~=~}& {\frac{1}{4}~+~\frac{9}{28}} &{} \\
{}&{}
& {~=~}& {\frac{4}{7}} &{} \\
\end{array}$
5. Suppose that an event F is: “getting exactly two heads”.
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Then we get: F = {HH}
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Now we use result 4 that we saw in the previous section:
$\rm{P(A)~=~\sum{P \left(\lbrace \omega_i \rbrace \right)}}$
•
So in our present case, we get:
$\begin{array}{ll}
{}&{\rm{P(F)}}
&
{~=~}& {\rm{P \left(\lbrace HH \rbrace \right)}} &{} \\
{}&{}
& {~=~}& {\frac{1}{4}} &{} \\
\end{array}$
The above two examples help us to understand how the probability of an event is calculated.
Probabilities of equally likely outcomes
This can be explained using an example. It can be written in 4 steps:
1. Consider the example of pens that we saw above.
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We saw that, all outcomes in S have the same probability.
2. We also saw that: P(A) = 3/8.
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Here the numerator ‘3’ is the number of elements in A. It is the number of outcomes favorable to A.
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Recall that, the number of elements in any set A is denoted as: n(A)
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So we can write:
When the outcomes in S are equally likely, the numerator of the probability for any event A is simply n(A)
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Now consider the denominator ‘8’. It is the number of elements in S.
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So we can write:
When the outcomes in S are equally likely, the denominator of the probability for event A is simply n(S)
3. Similarly, we saw that: P(B) = 1/2 = 4/8.
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Here the numerator ‘4’ is the number of elements in B. It is the number of outcomes favorable to B. It is n(B).
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Now consider the denominator ‘8’. It is n(S).
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Here also, we get the same result:
When the outcomes in S are equally likely, the denominator of the probability for event B is simply n(S)
4. So we can write it as a formula:
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When the outcomes in S are equally likely, the probability of any event A is given by:
$\rm{P(A)~=~\frac{n(A)}{n(S)}}$
The above formula is based on an example. We must write the general proof. It can be written in 3 steps:
1. Let S = {𝜔1, 𝜔2, 𝜔3, . . . , 𝜔k} be the sample space of an experiment.
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We see that, there are k outcomes. So n(S) = k
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Let us assume that, all those outcomes have the same probability. That is:
P({𝜔i}) = p, where 𝜔i ∈ S
2. Using axiom 2, we have:
For any experiment, P(S) = 1
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So for the present case, we can write:
P(S) = p + p + p + . . . + p
(Here p must be written k times because, number of elements in S is k)
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So we get: P(S) = kp = 1
Thus: $p = \frac{1}{k}$.
3. Let A be an event of the experiment. Let the set A have m elements. So n(A) = m.
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Then we can write:
P(A) = p + p + p + . . . + p
(Here p must be written m times because, number of elements in A is m)
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So we get: $\rm{P(A)~=~mp~=~m \times \frac{1}{k}}$
Thus: $\rm{P(A)~=~\frac{m}{k}~=~\frac{n(A)}{n(S)}}$
In the next section, we will see Probability of the event "A or B".
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