$P(B)$ for non-disjoint events when only $P(A)$ and $P(A \cap B)$ is given

Question

If I know $P(A)$ = 0.3 and $P(A \cap B)$ = 0.2, can I find minimum possible value of P(B)?

This is my thought process but it leads to negative probability. $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) \\ P(A \cup B) \geq 0 \\ 0.3 + P(B) - 0.2 \geq 0 \\ 0.1 + P(B) \geq 0 \\ P(B) \geq -0.1 $$

My other thought was that because we already know $P(A \cap B)$ = 0.2, the minimum value of $P(B)$ has to be 0.2.

I'm so lost though how to think about it. Thank you for your help!!

Answer 1

You are right that $P(B) \ge P(A \cap B)$.

In fact, $P(B)$ can be equal to $P(A \cap B)$ if we let $B = A \cap B$.

Hence, the minimal value is $0.2$.

Note that $P(B) \ge -0.1$ is just a bound.

Answer 2

Indeed the second method is the way to go.   Furthermore it gives the upper bound too.

Because: $0\leq \mathsf P(B\cap A^\complement)\leq \mathsf P(A^\complement)$, therefore: $$\begin{split}\mathsf P(A\cap B)&~\leq~ \mathsf P(B)&~\leq~\mathsf P(A\cap B)+\mathsf P(A^\complement)\\ 0.2&~\leq~ \mathsf P(B)&~\leq~ 0.9\end{split}$$

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Also, because $\mathsf P(A\cup B) ~{= \mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)\\ = \mathsf P(B)+0.1}$, therefore too, $0.3\leq \mathsf P(A\cup B)\leq 1.0$ .

Which is why the first method did not give the most lower bound.   You underestimated the inframum for the union's probability.