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Considering independent events/multiplictaion theorem:

In probability theory, to say that two events are independent means that the occurrence of one does not affect the probability of the other

Two events A and B are independent if and only if their joint probability equals the product of their probabilities:

$$P(A\cap B)=P(A)P(B)$$

Suppose this: $A=\xi$ and $B\subseteq A$ Then, $P(AB)=P(A)P(B)$, since $P(A)=1$, thus making them independent, but I can't see how actually they are independent. Suppose A doesn't happen then B can't happen so I see them dependent.

Another Case, let $S=\xi$ and $A,B\subseteq S$, consider $n(S)=10,n(A)=4,n(B)=5,n(AB)=2$, how can these be independent, because if one event doesn't happens or happens, the probability of happening of other is affected.

Why intutively, this seems wrong?

RE60K
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3 Answers3

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I can see how the first case might be confusing. However, it's a degenerate case in which $A$ can never not happen regardless of whether $B$ happens or not. Sure, $B$ can't happen without $A$ happening, but then again, nothing can.

In your second case, you're missing a concept known as conditional probability, in which you only take into account the probability that an event happens given that another event happens.

It's defined as follows: The probability of $A$ given $B$ (or $P(A|B)$) is the probability of both $A$ and $B$ happening divided by the probability of $B$ happening in total, or $\displaystyle \frac {P(AB)}{P(B)}$.

In your case, this is $\frac{2}{4}$ for $B$ occurring given that $A$ occurs as opposed to $\frac{5}{10}$ for $B$ occurring in the general case. Notice that these two fractions are the same (i.e. $P(B|A) = P(B)$) – the fact that $A$ occurred had no effect on the probability of $B$ occurring.

Joe Z.
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  • Basically, you were comparing the wrong things there $-$ you should be comparing $P(B|A)$ and $P(B)$, not $P(AB)$ and $P(B)$. – Joe Z. Nov 29 '14 at 05:32
  • why intutively this seems wrong? – RE60K Nov 29 '14 at 05:34
  • Probably because conditional probability isn't part of your intuition yet. It takes a bit of time to learn. – Joe Z. Nov 29 '14 at 05:40
  • yes it's been few days since I learnt that – RE60K Nov 29 '14 at 05:42
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    Reread my answer, and then answer the question: "Given that B has happened, is the probability of $A$ happening any different than the probability of $A$ happening in general?" If $B$ has happened, then we're no longer looking at the entire set $S$ to figure out whether $A$ has happened, only the 5 elements in $B$. So the probability becomes 2 out of 5, rather than 2 out of 10. You can't just compare raw set sizes; probability is always a ratio. – Joe Z. Nov 29 '14 at 05:47
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Suppose this: A=ξ and B⊆A Then, P(AB)=P(A)P(B), since P(A)=1, thus making them independent, but I can't see how actually they are independent. Suppose A doesn't happen then B can't happen so I see them dependent.

If $\mathsf P(A)=1$ then $A$ can never not happen.   This is independent of the occurance of $B$.

A similar case is when $A=\varnothing$.   Then $\mathsf P(A,B)=\mathsf P(A)\mathsf P(B)=0$.   This is because $A$ can almost surely never happen, independent of the occurence of $B$.

Another Case, let S=ξ and A,B⊆S, consider n(S)=10,n(A)=4,n(B)=5,n(AB)=2, how can these be independent, because if one event doesn't happens or happens, the probability of happening of other is affected.

Not so.   $B$ happens half of the time, $B$ happens half of the times than $A$ happens, and $B$ happens half of the times $A$ doesn't happen.   $A$ happens $2/5$ of the time, $A$ happens $2/5$ of the times $B$ happens, and $A$ happens $2/5$ the times $B$ doesn't happen. $$\boxed{\begin{array}{c|c}A & A & AB & AB & B & B & B & \_ & \_ & \_ \end{array}}$$

$$\frac{n(A)}{n(S)}=\frac{n(AB)}{n(B)}=\frac 2 5, \quad \frac{n(B)}{n(S)}=\frac{n(AB)}{n(A)}=\frac 1 2\\\frac{n(A)}{n(S)}\frac{n(B)}{n(S)}=\frac{n(A)}{n(S)}=\frac{1}{5}$$

Graham Kemp
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The independence is given by the formal definition, hence in both cases the events are independent. But I understand, that you are expecting some intuitions.

In the first case $A$ happens with probability one, hence it is impossible (in probability sense) that it doesn't happen, as you are suggesting. In the second case, let's imagine ten squares and sets $A$, $B$, $A\cap B$ consisting in the proper way of 4, 5 and 2 squares, respectively. Let now in the same moment we have two shots such that each of them must target one of 10 squares with the equal probability. How probabilities could be affected? (We do not consider collision of bullets).

Przemysław Scherwentke
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