Given: p(A) = 0.6 and p(B) = 0.4 and p(AB) = 0.2
What does p(AB) mean, and how is it different from p(A U B)? How would I find p(A U B) with the given information?
How do I calculate p(AcBc)?
$\endgroup$2 Answers
$\begingroup$$P(AB)$ generally means $P(A\cap B)$ and by inclusion-exclusion principle $$P(A\cup B) = P(A) + P(B) - P(AB)$$
$P(A^c)$ means the probability that $A$ doesnt happen.
Since we have $A\cup A^c = \Omega$ then $P(A\cap A^c) = 1$ and then, by the above law (noting $A\cap A^c =\emptyset$) we have
$$P(A^c) = 1 - P(A)$$
And similary $P(A^CB) = P(A^c \cap B)$ and since $B = (B\cap A)\cup (B\cap A^C)$ we have $$P(B) = P(A^CB) + P(AB)$$
And then $P(A^CB) = P(B) - P(AB)$
Using the same idea, $P(A^CB^C) = P( (A\cup B)^C) = 1-P(A\cup B) = 1 + P(AB) - P(A) - P(B)$
$\endgroup$ $\begingroup$$P(AB)$ means the probability that events $A$ and $B$ occur. You could write it $P(A\cap B)$. The superscript $c$ means "complement" and $A^c$ means all outcomes not in $A$. So, $P(A^cB)$ means the probability that not-$A$ and $B$ both occur, etc.
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