Maximum and minimum values of probabilities

Question

If $P\left(A\right)=0.8\:$ and $P\left(B\right)=0.4$, find the maximum and minimum values of $\:P(A|B)$.

My textbook says the answer is $0.5$ to $1$. But I think the answer should be $0$ to $1$.

The textbook claims $P(A∩B)$ is $0.2$ when $P(A'∩B')=0$

I think that the minimum value arises when $A$ and $B$ are mutually exclusive. So there isn't a chance of both happening and so you have $\frac{0}{0.4} = P(A|B) = 0$ (right?)

I agree with the textbook in saying the maximum value is one.

Answer 1

I think a visualization helps most with understanding these probability bounds.

To begin, $$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{P(A \cap B)}{0.4}\text{.}$$ To finish this problem, we need to find bounds for $P(A \cap B)$. Let's draw a Venn diagram for these two events. Notice that $P(A) = x+y$, $P(B) = y+z$, and $P(A \cap B) = y$.

Since $P(A) = 0.8$ and $P(B) = 0.4$, we know that $$\begin{align} x+y &= 0.8 \\ y+z &= 0.4\text{.} \end{align}$$ Furthermore, since $x$, $y$, and $z$ all are greater than or equal to $0$, we know immediately that $y$ must be less than or equal to $0.4$ by the equations above. ($y$ can't exceed $0.4$, otherwise $y+z$ would exceed $0.4$, which is a contradiction with the above system of equations.)

Lastly, we have to find the smallest value of $y$. $y$ cannot be $0$, because otherwise $x + z = 1.2$, and one cannot have probabilities summing to exceed $1$ in a sample space.

So, the way I would approach this is as follows: we know $P(A)$ and $P(B)$. What's the largest value that I can subtract from their sum $P(A) + P(B)$ so that I get probabilities summing up to $1$?

Well, we know that $$0.8 + 0.4 - 0.2 = 1$$ so we would be subtracting $0.2$. This amounts to $0.2$ being the smallest possible value of $y$, hence $$0.2 \leq y = P(A \cap B) \leq 0.4$$ or $$\dfrac{0.2}{0.4} = \dfrac{1}{2} \leq \dfrac{P(A \cap B)}{P(B)} \leq \dfrac{0.4}{0.4} = 1$$ or $$\dfrac{1}{2} \leq P(A \mid B) \leq 1\text{.}$$

Answer 2

Since $P(A\cap B)$ is less than both $P(A)$ and $P(B)$,

we have $$P(A\cap B)\le \min(P(A),P(B))=0.4$$ And from Bonferroni's inequality it follows that $$P(A\cap B)\ge \max(0,P(A)+P(B)-1)=0.2$$

Now, $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{0.4}$$

So, $$0.5\le P(A\mid B)\le 1$$

Answer 3

For completeness, I'll show both bounds of $P(A|B)$.

Lower Bound

We note that $$P(A)+P(B)-P(A\cup B) = P(A\cap B)$$ which is a simple rearrangement of the standard probability summation formula. Note that the upper bound of $P(A\cup B)=1$, so when we substitute, we get $0.8+0.4-1=P(A\cap B) = 0.2$

Hence, $$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac {0.2}{0.4} = \frac12$$

Upper Bound

We reach an upper bound when $A\rightarrow B$. So, $P(A\cap B)=P(B)$, so $$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$$

Intuition for Lower Bound

So why does the formula used to compute the lower bound work? Well, that is because when we want to count the probability of $P(A\cup B)$, we realize that its the sum of $P(A)+P(B)$, but we have double counted the probability where they overlap, notably $P(A\cap B)$. This implies that $$P(A)+P(B)-P(A\cap B) = P(A\cup B)$$