Let $A, B$ be events with probabilities $P(A)=2/5$, $P(B)=5/6$, respectively. Find the best lower and upper bound of the probability $P(A \cap B)$ of the intersection $A \cap B$. Namely, find real numbers $a, b$ such that
\[a \leq P(A \cap B) \leq b\] and $P(A \cap B)$ could take any values between $a$ and $b$.

Solution.

For the lower bound, we use the basic equality of the probability theory that
\[P(A \cap B) = P(A) + P(B) – P(A \cup B).\] Note also that we always have the inequality $P(A \cup B) \leq 1$. Altogether we obtain the following inequality.
\begin{align*}
P(A \cap B) &= P(A) + P(B) – P(A \cup B)\\
&\geq P(A) + P(B) – 1\\[5pt] &= \frac{2}{5} + \frac{5}{6} – 1\\[5pt] &= \frac{12 + 25 -30}{30}\\[5pt] &= \frac{7}{30}.
\end{align*}

This gives the lower bound $a = 7/30$. Note that $P(A \cap B)$ could take this lower bound when $P(A \cup B) = 1$ and this happens if $A\cup B$ is the whole sample space.

Next, we’ll obtain the upper bound. As the intersection $A \cap B$ is contained in the set $A$ and in the set $B$, we have
\begin{align*}
P(A \cap B) &\leq \min(P(A), P(B))\\
&= \min \left(\frac{2}{5}, \frac{5}{6}\right)\\
& = \frac{2}{5}.
\end{align*}

This yields the upper bound $b = 2/5$. The probability $P(A \cap B)$ could take this upper bound when $A \cap B = A$ (this happens when $A \subset B$).

In conclusion, we obtain the following bounds
\[\frac{7}{30} \leq P(A \cap B) \leq \frac{2}{5}.\]

We remark that as a probability we clearly have bounds $0 \leq P(A \cap B) \leq 1$. However, these bounds are not optimal in a sense that $P(A \cap B)$ never takes values less than $7/30$ or above $2/5$ as we determined above.

The post Lower and Upper Bounds of the Probability of the Intersection of Two Events first appeared on Problems in Mathematics.