Let A, B, and C be events such that A and B are both subsets of C. Also, let P(A) = 0.3, P(B) = 0.4, and P(C) = 0.6. Then, P(A|B) could be what

Question

Let A, B, and C be events such that A and B are both subsets of C. Also, let P(A) = 0.3, P(B) = 0.4, and P(C) = 0.6. Then, P(A|B) could be what, there is hint saying that there is not only one answer and I do not get it.

Answer 1

Well, if $A$ and $B$ are independent, then $P(A \vert B) = P(A) = 0.3$.

If $A$ and $B$ are dependent, then $$P(A \vert B) = \dfrac{P(A \cap B)}{P(B)}.$$

And then it can depend again:

• if $A \cup B = C$, then $P(C) = P(A) + P(B) - P(A \cap B)$, so $P(A \cap B) = 0.3 + 0.4 - 0.6 = 0.1$. thus $P(A \vert B ) = \frac{1}{4} = 0.25$.

• if $A \cup B\neq C$, then there exists $D \subseteq C$ such that $A \cup B \cup D = C$, with $D \cap (A \cup B) = \emptyset$. In this case, we have $$P(C) = P(A \cup B ) + P(D) = P(A) + P(B) - P(A \cap B) + P(D)$$ and $P(A \cap B)$ can be "anything", depending on $D$.

Answer 2

$$0.6\geq P(A\cup B)=0.3+0.4-P(A\cap B).$$

Thus, $$P\left(A|B\right)=\frac{P(A\cap B)}{P(B)}\geq\frac{0.1}{0.4}\geq\frac{1}{4}.$$ Also, use that $P(A\cap B)\leq P(A).$

Answer 3

$$P(A\cup B)\geq \max{\{P(A),P(B)\}}=0.4$$ $$P(A\cup B)\leq \min{\{P(C),P(A)+P(B)\}}=0.6$$ $$P(A\cup B)\in[0.4,0.6]$$ $$P(AB)=P(A)+P(B)-P(A\cup B)\in[0.1,0.3]$$ $$P(A|B)=\frac{P(AB)}{P(B)}\in[0.25,0.75]$$