For a toss of fair die, if events are, A: {1,2}, B: {2,4,6}, and C: {4,5,6}, then A and B are independent but B and C are not. Why?

Question

As far as I understood independence, A and B should not be independent since if either of them happens then we can tell something about the other one. But if they are independent then B and C should also be. From mathematical proof it is explainable, but I didn't understand the intuition behind it.

The example is from “Elementary Bayesian Statistics” by Gordon Antelman, Chapter 2

Text:

For a fair die $U = (1,2,3,4,5,6)$. Let three events be defined as:

• $A \equiv (1,2)$, so $P(A)=2/6$

• $B \equiv (2,4,6)$, so $P(B)=3/6$, and

• $C \equiv (4,5,6)$, so $P(A)=3/6$

Then

• $A \cap B = (2)$ and $P(A \cap B) = 1/6 = P(A)P(B) = (1/3)(1/2)$, so A and B are independent.

• $B \cap C = (4,6)$ and $P(B \cap C) = 2/6 \ne P(B)P(C) = (1/2)(1/2)$, so B and C are not independent.

Answer 1

Other explanation why $\color{red}{A = \{1, 2\}} $ and $\color{blue}{B = \{2, 4, 6\}}$ are independent:

Someone rolled a die:

• The probability of the event $\color{red}A$ is $\color{red}{2\mkern-0.1ex/6}$, i.e. $\color{red}{1\mkern-0.1ex/3}$.
• Someone tells you that the result is an $\color{blue}{\text{even}}$ number (the event $\color{blue}B$). In spite of this new info the probability of $\color{red}A$ is still $\color{red}{1\mkern-0.1ex/3}$
  ($\color{blue}2 \in \color{red}A$, but $\color{blue}{4} \notin \color{red}A$, $\color{blue}6 \notin \color{red}A$).

In the opposite way —

• The probability of event $\color{blue}B$ is $\color{blue}{3\mkern-0.1ex/6}$, .i.e. $\color{blue}{1\mkern-0.1ex/2}$.
• Someone tells you that the result is a number $\color{red}1$ or $\color{red}2$ (the event $\color{red}A$). In spite of this new info the probability of $\color{blue}B$ is still $\color{blue}{1\mkern-0.1ex/2}$
  ($\color{red}2 \in \color{blue}B$, but $\color{red}{1} \notin \color{blue}B$).

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Note:

You may use the same scheme to see that the events $\color{blue}B$ and $\color{green}C$ are not independent.

Answer 2

You probably confuse indepenent events for mutually exclusive events.

... A and B should not be independent since if either of them happens then we can tell something about the other one.

It is not true. What does mean “something”? The probability of “the other one” didn't change!

The correct formulation should be

• “A and B are independent — if either of them happens, then it doesn't change the probability of the other”.

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As I wrote, you are probably confused mutually exclusiveness (the empty intersection, distinctiveness) for the independence of two events.

But they are two independent things:

• the events $\{1\}$ and $\{2, 4, 6\}$ are mutually exclusive, but they are not independent,

• the events $\{2\}$ and $\{2, 4, 6\}$ are neither mutually exclusive, nor independent,

• the events $\{1, 2\}$ and $\{2, 4, 6\}$ are not mutually exclusive, but they are independent,

• the events $\emptyset$ and $\{2, 4, 6\}$ are mutually exclusive, and they are independent.

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Independence can arise in two distinct ways:

• we explicitly assume that events are independent (e.g. rolling a die again), or
• we derive independence by verifying that it fulfills the formula $P(A \cap B) = P(A)P(B)$.

Generally, there is no way to “see” the independence (e.g. by looking in a Venn diagram).

By other words, if we may “see” the independence of two events, then they certainly fulfill the formula. But the opposite is not true — the formula is fulfilled, the independence is guaranteed, but there is not a visible reason, why.

The idea of independent events arises from the evident, noticable independence, but is not limited to it.