Let's say we flip a coin twice:

The sample space (set of all possible outcomes) is:

$$\Omega = \{HH, HT, TH, TT\}$$

Now let's say we have the following two events (sets of outcomes):

$$A= \{HH, HT, TH\}, B = \{HT, TH, TT\}$$ where $A$ is the event that at least one coin is heads and $B$ is the event that at least one coin is tails.

Now the intersection between A and B ($A\cap B$) is the set of outcomes that are shared between A and B, the event (C) where at least one coin is heads AND at least one coin is tails:

$$C = A\cap B = AB = \{HT, TH\}$$

In this trivial example, I can calculate the probability of C by taking the size of the event over the size of the sample space: $$P(C) = \frac{size(C)}{size(\Omega)} = \frac{2}{4} = 0.5$$

But I am struggling to understand how I would find this answer if the sets were too large or complicated to count out.

The calculation $P(A)*P(B) = \frac{9}{16}$ clearly gives a different answer. I think this means that events A and B are not independent. Two events are dependent if the outcome of the first event affects the outcome of the second event, so that the probability is changed. So conceptually, how/why is A affecting B (or vice-versa)?

And more generally, if it's not conceptually clear that one event is influencing another, does that mean the only way to determine if they are independent is by collecting data from observations?

  • $\begingroup$ $P(A \cup B) = P(A)*P(B)$ is only true for independent events. You could potentially use that $\endgroup$
    – ENV
    Mar 10, 2021 at 3:51
  • $\begingroup$ "Why are they not independent" Because the calculation $\Pr(A)\times \Pr(B)$ gives a different result than $\Pr(A\cap B)$, plain and simple. You could also consider checking if $\Pr(A\mid B)=\Pr(A)$, checking if $\Pr(A\mid B^c)=\Pr(A)$ and so on... Here, for your example, it should be plain to see that if it is not the case that there is at least one tail, i.e. that $B^c$ is true, then it must be the case that all coins came up heads implying that $A$ is true. That is to say, $\Pr(A\mid B^c)=1$. Meanwhile it is plain to see that $\Pr(A)\neq 1$, thus implying dependence. $\endgroup$
    – JMoravitz
    Mar 10, 2021 at 3:52
  • $\begingroup$ Now... we could go and talk about ley ways of thinking of independence, one thing not affecting another because of being unrelated events which don't involve the same processes... such as two separate coins being flipped each being their own physical object not affecting the spin of the other as they are flipped... but that misses out on recognizing that some events despite involving the same objects might still be independent, e.g. the event the first card in a deck is a king and the event the second card in a deck is a heart... or first coin is heads vs odd number of heads. $\endgroup$
    – JMoravitz
    Mar 10, 2021 at 3:57

1 Answer 1


I have a potential answer to my question. I think the issue I am having is conceptualizing the definition of "dependent events." I think this is a more useful definition: Events are dependent if the outcome of one affects the probability that the conditions of the other event will be satisfied (AKA, that it would occur).

In the example in the question, we can see that the outcomes of event A will indeed affect the probability that event B will occur. If the outcomes of A are either HT or TH, then the probability of B occurring is 100%. That is because the conditions of B, at least one Tail, is satisfied by both those outcomes. If the outcome of A is HH, then the probability of B occuring is 0%. Again, because the conditions of B, at least one Tail, is not met by that outcome.


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