I have a universe, $ U = \{a, b, c, d, e, f\}$ and sets $A = \{a, b, c\} and B = \{a, d, e, f\}$

If $P(A) = P(X = x \in A)$ and $P(B) = P(X = x \in B)$, where $X$ is a random variable defined by uniformly selecting elements of $U$.

Are these values of unconditional, joint and conditional probabilities correct?

$ p(A) = 1/2$

$p(A,B) = 1/6$

$P(A|B) = 1/4$

Thanks in advance.


If you are assuming that each object of your universe has the same probability then each object has $P(x=x)=\frac{1}{6}$. So as $a,b$ and $c$ have the same probability and we are assuming they are independent, then $P(x\in A) = P(x=a)+P(x=b)+P(x=c) = \frac{3}{6} = \frac{1}{2}$. Then, $P(x \in A \cap B) = P(x = a) = \frac{1}{6}$ because $A \cap B = a$. Finally, $P(A|B)$ is the probability of $a$ in $B$ universe, so here, as ther are 4 objects with the same probability then each one has $P(x=x)=\frac{1}{4}$, so $P(x \in A)=P(x \in A\cap B)=P(x=a) = \frac{1}{4}$.

Another way is using the conditional formula $P(A|B)=\frac{P(A\cap B)}{P(B)} = \frac{P(x=a)}{P(x\in B)} = \frac{\frac{1}{6}}{\frac{4}{6}} = \frac{1}{4}$. So yes, your answer where correct.

  • $\begingroup$ Thanks. Superb explanation. $\endgroup$ – user102859 Feb 10 at 10:30
  • $\begingroup$ @JoseSquare, please could you help me by writing an example where no uniform distribution is assumed? Thanks. $\endgroup$ – Nacho Sep 6 at 16:03

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