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If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

(a):

Because we don't have a lot of events, we won't make the three diagram. Using

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space}}}}\)

we have that

\(\begin{aligned}P(A) &=& \frac{{\# {\rm{ of favorable outcomes in A}}}}{{\# {\rm{ of outcomes in the sample space}}}} &=& \frac{{2000}}{{10,000}}\\ &=& 0.2.\end{aligned}\)

The following is true

\(\begin{align}\text P(B)=&P \!\![\!\!\text{ (A}\cap B)\cup ({A}'\cap B)]......(1) \\ \text =&P(A\cap B)+P({A}'\cap B)......(2) \\ \text =&P(A)*P(B \!\!|\!\!\text{ A)+P(B }\!\!|\!\!\text{ }{A}')P({A}')......(3) \\ \text =&0 \text{.2*}\frac{\text{1999}}{\text{9999}}\text{+(1-0}\text{.2)*}\frac{\text{2000}}{\text{9999}}......(4) \\ \text =&0 \text{.2}\text{.}\end{align}\)

(1): this refers to any of the events A and B;

(2) due to the fact that they are separate events;

(3): we might have used Bayes' Theorem, but we'll apply the multiplication rule below instead.

(4): the conditional probability, using

\({\rm{P(B|A) = }}\frac{{{\rm{\# of outcomes in the sample spaceB|A}}}}{{{\rm{\# of outcomes in the sample space'}}}}\)

is the year \({\rm{1999/9999}}\)Because \({\rm{1999}}\) is the number of positive occurrences (we chose one out of \({\rm{2000}}\) in the first round), and \({\rm{9999}}\) is the total number of outcomes available, the result is 1999/9999. (because we took one out of \({\rm{10,000}}\)). The probability of \({\rm{P}}\left( {{\rm{B|}}A'} \right)\).

The Rule of Multiplication

\(P(A \cap B) = {\rm{P(A|B)*P(B) }}\)

The likelihood of the events colliding is

\(\begin{array}{c}P(A \cap B) &=& {\mathop{\rm P}\nolimits} (A)*P(B|A)........(1)\\ \rm &=& 0{\rm{.2*}}\frac{{{\rm{1999}}}}{{{\rm{9999}}}}.......(2)\\ \rm &=& 0{\rm{.03998}}\end{array}\)

(1): (1) by employing the multiplication rule;

(2): see the preceding description.

Two occurrences A and B are independent if and only if they have the same multiplication property.

\(P(A \cap B){\rm{ = P(A)*P(B) }}\)

We can notice that

\(\begin{aligned}{l}P(A \cap B) =& {\rm{0}}{\rm{.03998/0}}{\rm{.04}}\\ \rm =& 0{\rm{.2*0}}{\rm{.2 = P(A)*P(B)}}\end{aligned}\)

therefore, the events are dependent.

(b):Because we know that A and B are independent, we can use the multiplication property to prove the following.

\(\begin{aligned}{c} \rm =& P(A)*P(B) \\ \rm =& 0{\rm{.2*0}}{\rm{.2}}\\ \rm =& 0{\rm{.04}}{\rm{.}}\end{aligned}\)

The difference between probabilities of intersection in (a) and (b) is very small,\(0.0002.\)

Practically, we can assume that the events are independent, because the sample size is big. We would obtain essentially

Therefore,the correct probability for big samples \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right){\rm{ = 0}}{\rm{.04}}\)

The entire sample size is now ten, and the probability are as follows:\({\rm{P}}\left( {\rm{A}} \right){\rm{ = P}}\left( {\rm{B}} \right){\rm{ = 0}}{\rm{.2}}{\rm{.}}\)

We have the probability

\({\rm{P(A)*P(B) = 0}}{\rm{.04}}{\rm{.}}\)

However,

\(\begin{array}{l}P(A \cap B) &=& {\rm{P(A) * P(B|A)}}.........{\rm{(1)}}\\ \rm &=& \frac{{\rm{2}}}{{{\rm{10}}}}*\frac{{\rm{1}}}{{\rm{9}}}\\ \rm &=& 0{\rm{.0222}}\end{array}\)

(1):from the multiplication rule given above

We can see that the probabilities differ a lot now (0.0222 and 0.04), therefore it does not return about the proper result for the intersection. This is due to the fact that the sample size for case (a) is 1000 times larger!

As previously stated, when the sample size is large, the assumption of independence is true for calculating the probability of intersection.

Therefore , it does not independence