1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
3
$\begingroup$

$A,B,C$ 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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.