0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.