I have the following problem which I'm not sure how to solve:

Let $\Omega=\{A,B,C\}$ be the sample space of events happening on some experiment and $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$. Find the probabilities of all random events related with this experiment.

My understanding is that I have to find all subsets of $\Omega$, but I'm not sure how to do that since the events might be dependent on each other.

I try $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ but I don't know whether $P(A\cap B)=P(A)P(B)$.

Another approach I try is:

$P(\overline A)=P((B\cup C) \setminus A)=\frac{1}{2}$ $P(\overline B)=P((A\cup C) \setminus B)=\frac{2}{3}$,

but I don't think this takes me anywhere too.

Is there a missing condition in the statement of the problem for independence of the events, or is there a way to solve the problem which I'm missing.

Thanks in advance!

  • $\begingroup$ Yes, Thank you! A and B are elements of the sample space and not sets. Therefore their intersection is the empty set. $\endgroup$ – Nikola Aug 28 '17 at 16:11
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    $\begingroup$ @NikolaShahpazov Not necessarily, one could choose a probability space made of sets. :-) $\endgroup$ – Did Aug 28 '17 at 16:12
  • $\begingroup$ @mfl Yes. Your point being? $\endgroup$ – Did Aug 28 '17 at 17:33

$A,B,C$ aren't events. They're sample points.

Events are subsets of the sample space.

By definition, for a finite sample space, the probability of an event is the sum of the probabilities of the sample points which are elements of the event.

Also, for a finite (or countable) sample space, the sum of the probabilities of all the sample points must be $1$, so $C$ has probability $1/6$.

Since the sample space has $3$ elements, it has$\;2^3=8\;$subsets. Thus, there are$\;8\;$events. For each event, write it in set notation (for example, $\{A,B\}$), and find its probability by summing the probabilities of its sample points.

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    $\begingroup$ Yes! Thank you! So A, B and C are outcomes and not events. Maybe what got me confused is that they are denoted by the letters which are most used for naming events :). I'm used to seeing outcomes denoted $\omega_1, \omega_2,...$. I'm still training myself in not getting confused by different notations. $\endgroup$ – Nikola Aug 28 '17 at 16:09

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