Q.State whether the following statement is True or False: If $A$ and $B$ are two events such that $P(A) > 0$ and $P(A) + P(B) > 1$, then $P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}$.

Concept: Probability Complement Rule — rewriting $P(B')$ as $1 - P(B)$ and checking the inequality.

We are given $P(A) > 0$ and $P(A) + P(B) > 1$.

The statement to check is:

Since $P(B') = 1 - P(B)$, the right-hand side becomes:

Now $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$. The inequality is therefore:

Multiplying through by $P(A) > 0$ gives:

$P(A \cap B) \ge P(A) + P(B) - 1.$ …

Why This Approach Works

The problem asks whether $P(B \mid A) \ge 1 - \dfrac{P(B')}{P(A)}$ is always true under the conditions $P(A) > 0$ and $P(A) + P(B) > 1$.

At first glance, this looks like a conditional probability inequality that might depend on the specific events. But the complement rule gives us a powerful way to simplify: $P(B') = 1 - P(B)$. So the right-hand side becomes $1 - \dfrac{1 - P(B)}{P(A)}$.

The trick is to realise that $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$, and we can relate $P(A \cap B)$ to $P(A) + P(B) - 1$ using the inclusion-exclusion principle. The condition $P(A) + P(B) > 1$ guarantees that $P(A \cap B) > 0$, which is crucial.

Let's work through it step by step.

Step-by-Step Solution

1. Write the target inequality in terms of $P(A \cap B)$.

We know $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$. The inequality becomes:

Multiply both sides by $P(A) > 0$ (so the inequality direction stays the same):

2. Replace $P(B')$ using the complement rule.

Since $P(B') = 1 - P(B)$, we get:

So the inequality we need to prove is:

3. Recognise this as a known inequality from inclusion-exclusion.

The inclusion-exclusion principle for two events states:

Since $P(A \cup B) \le 1$ (probabilities cannot exceed 1), we have:

Rearranging:

This is exactly the inequality we need! It holds for any two events $A$ and $B$, regardless of the given conditions.

4. Check the role of the given conditions.

• $P(A) > 0$: This is necessary so that $P(B \mid A)$ is defined (we can't divide by zero). …

Method: Verifying a conditional-probability inequality

To test an inequality in $P(B\mid A)$, clear the conditional into $P(A\cap B)$ and reduce to a known probability bound.

Steps

Step 1: Replace the conditional probability.

Multiply the whole inequality by $P(A)>0$ (direction unchanged, since $P(A)>0$).

Step 2: Use the complement rule to simplify the other side.

Wherever $P(B')$ appears, write $P(B')=1-P(B)$ and collect terms.

Step 3: Recognise the resulting inequality.

The target typically reduces to

$P(A\cap B)\ge P(A)+P(B)-1,$ …

Common Mistakes

Mistake 1: Reversing the inequality when clearing $P(A)$.

Why it's wrong: multiplying by $P(A)$ keeps the direction only because $P(A)>0$; treating it as if it could flip is an error. Correct approach: since $P(A)>0$, the direction is unchanged, reducing to $P(A\cap B)\ge P(A)+P(B)-1$.

Mistake 2: Thinking the bound needs the condition $P(A)+P(B)>1$. …