Q.Fill in the blank: If $A$ and $B$ are such that $P(A' \cup B') = \dfrac{2}{3}$ and $P(A \cup B) = \dfrac{5}{9}$, then $P(A') + P(B') =$ __________.

The key idea is the Probability Complement Rule: $A' \cup B' = (A \cap B)'$ by De Morgan’s law.

Step 1:

$P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B) = \frac{2}{3}$.

Thus $P(A \cap B) = 1 - \frac{2}{3} = \frac{1}{3}$.

Step 2:

We know $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

So $\frac{5}{9} = P(A) + P(B) - \frac{1}{3}$.

Step 3: …

Let’s unpack why this works. The problem gives us two probabilities: one for the union of complements, and one for the union of the original events. At first glance, these might seem unrelated, but the complement rule ties them together beautifully.

The complement of $A' \cup B'$ is $(A' \cup B')' = A \cap B$. So $P(A' \cup B') = 1 - P(A \cap B)$. This is the crucial bridge — it lets us find $P(A \cap B)$ directly.

Now, we also have $P(A \cup B)$. The standard formula for the union is:

We don’t know $P(A)$ or $P(B)$ individually, but we don’t need them — we need $P(A') + P(B')$, which is $[1 - P(A)] + [1 - P(B)] = 2 - [P(A) + P(B)]$.

So if we can find $P(A) + P(B)$, we’re done. And that’s exactly what the union formula gives us, once we know $P(A \cap B)$.

Let’s go step by step.

1. Find $P(A \cap B)$ from the complement union. We have $P(A' \cup B') = \dfrac{2}{3}$. Since $A' \cup B' = (A \cap B)'$, we get:

Therefore:

2. Use the union formula to relate $P(A) + P(B)$ and $P(A \cap B)$. We know $P(A \cup B) = \dfrac{5}{9}$. So:

Solve for $P(A) + P(B)$: …

Method: De Morgan's laws with the complement rule

Use this when a problem mixes complements with unions/intersections — e.g. it gives $P(A'\cup B')$ and asks about $A$, $B$, or their complements.

Steps

Step 1: Convert the complement-of-a-combination using De Morgan.

Choosing the right one is essential — the union of complements is the complement of the intersection, not of the union.

Step 2: Apply the complement rule to get a probability.

$P\big((A\cap B)'\big) = 1 - P(A\cap B) \ \Rightarrow\ P(A\cap B) = 1 - P(A'\cup B').$ …

Common Mistakes

Mistake 1: Assuming $P(A'\cup B')=1-P(A\cup B)$.

Why it's wrong: the complement of $A\cup B$ is $A'\cap B'$ (intersection), not $A'\cup B'$. By De Morgan, $A'\cup B'=(A\cap B)'$. Correct approach: $P(A'\cup B')=1-P(A\cap B)$, so $P(A\cap B)=1-\tfrac23=\tfrac13$.

Mistake 2: Trying to find $P(A)$ and $P(B)$ individually. …