5.5 Independence

Two events are said to be independent if \(Pr(A\cap B)=Pr(A) \cdot Pr(B)\). The events \(A\) and \(B\) are independent if knowledge of \(B\) does not affect the probability of \(A\). This can be written in terms of conditional probability as \[Pr(A|B)=Pr(A)\\ Pr(B|A)=Pr(B)\] The probability of both events should be positive because division by zero is meaningless. Remember that \(Pr(A|B)=Pr(A\cap B)/Pr(B)\) and \(Pr(B|A)=Pr(A\cap B)/Pr(B)\).

Suppose you are rolling a red die and a green die. Let event \(A\) be “4 on the red die” and event \(B\) is “sum of the dice is odd”. What is \(Pr(A)\), \(Pr(B)\), and \(Pr(A \cap B)\)? Are \(A\) and \(B\) independent? The table below displays all possible outcomes and can help to calculate the probabilities.

Red 1 2 3 4 5 6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 5,3 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6

The probabilities are as follows: \[Pr(A)=6/36=1/6\\ Pr(B)=18/36 = 1/2\\ Pr(A\cap B) = 3/36 = 1/12\] To check for independence, multiply \(Pr(A)\) and \(Pr(B)\) as follows \[Pr(A) \cdot Pr(B)=\left(\frac{1}{6}\right)\left(\frac{1}{2}\right)=\frac{1}{12}\] Next consider a different scenario. Assume event \(C\) be “at least three dots on a die” and event \(D\) as “sum of dice equals 7”. Are those two events independent? \[Pr(C) = \frac{32}{36}\\ Pr(D) = \frac{1}{6}\]

\[Pr(C|D) = \frac{Pr(C \cap D)}{Pr(D)}=\frac{6/36}{6/36}=1\] Thus, the two events are dependent.

5.5.1 Birthday Problem

Usually, people underestimate the probability of finding two matching birthdays in a group of people. Here the issue is that people think in terms of matching with their own birthday. That probability is indeed small (1 out of 365). But the matching birthday considers all combinations. Define the following two events: Event \(B_2\) is that “two people have different birthdays” and event \(B_3\) is that “three different birthdays of three people.” The probability that two people have different birthdays is \[Pr(B_2) = 1-\frac{1}{365}\] and that probability that three different people have different birthdays is given by \[Pr(B_3) = Pr(A_3|B_2) \cdot Pr(B_2)\] \[Pr(A_3|B_2) = 1-\frac{2}{365}\] \[Pr(B_3) = \left(1-\frac{2}{365}\right)\left(1-\frac{1}{365}\right)\] In general, we have \[Pr(B_n) =Pr(A_n|B_{n-1}) \cdot Pr(B_{n-1}) = \left(1-\frac{n-1}{365}\right) \cdot Pr(B_{n-1})= \left(1-\frac{n-1}{365}\right) \cdots \left(1-\frac{2}{365}\right)\left(1-\frac{1}{365}\right)\] This last probability does to zero very quickly as \(n\) increases. There are also other examples of paradoxes of probability and other statistical strangeness.