Q.Evaluate $P(A \cup B)$, if $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$.

Concept: Conditional Probability — we use $P(A|B) = \frac{P(A \cap B)}{P(B)}$ to find the intersection, then the addition rule.

Step 1: From $2P(A) = P(B) = \frac{5}{13}$, we get

$P(B) = \frac{5}{13}$ and $P(A) = \frac{1}{2} \cdot \frac{5}{13} = \frac{5}{26}$.

Step 2: Using $P(A|B) = \frac{2}{5}$,

$P(A \cap B) = P(A|B) \cdot P(B) = \frac{2}{5} \cdot \frac{5}{13} = \frac{2}{13}$.

Step 3: By the addition rule, …

The problem gives you two probabilities and a conditional probability, and asks for the probability of the union. The natural instinct is to reach for the addition rule:

We already have $P(B)$ directly. But $P(A)$ is given indirectly through $2P(A) = P(B)$, and $P(A \cap B)$ is hidden inside the conditional probability $P(A|B)$. So the plan is: first extract $P(A)$ and $P(A \cap B)$, then plug into the union formula.

1. Find $P(A)$ from the given relation

We know $2P(A) = P(B) = \frac{5}{13}$.

So $P(B) = \frac{5}{13}$, and therefore

2. Find $P(A \cap B)$ using the definition of conditional probability

The definition:

We are given $P(A|B) = \frac{2}{5}$ and $P(B) = \frac{5}{13}$. Substitute:

Multiply both sides by $\frac{5}{13}$:

$P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$ …

Method: Decoding indirect givens, then the addition rule

Use this when the union $P(A\cup B)$ is wanted but $P(A)$ and $P(A\cap B)$ are hidden inside relations or a conditional probability rather than stated outright.

Steps

Step 1: Unpack the single-event probabilities

A relation such as $mP(A)=P(B)=\text{value}$ pins down each individual probability: read $P(B)$ directly, then solve the small equation for $P(A)$.

Step 2: Convert the conditional into the intersection

Apply the multiplication rule to the given conditional:

This turns a "given that" statement into the joint probability the union formula needs. …

Common Mistakes

Mistake 1: Misreading $2P(A)=\tfrac{5}{13}$ as $P(A)=2\times\tfrac{5}{13}=\tfrac{10}{13}$.

Why it's wrong: $2P(A)=\tfrac{5}{13}$ means $P(A)=\tfrac12\cdot\tfrac{5}{13}=\tfrac{5}{26}$ — you divide by 2, not multiply. The value $\tfrac{10}{13}$ also breaks the later union.

Mistake 2: Computing $P(A\cap B)$ as $P(A\mid B)\,P(A)$ instead of $P(A\mid B)\,P(B)$.

Why it's wrong: since $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$, we get $P(A\cap B)=P(A\mid B)\,P(B)=\tfrac25\cdot\tfrac{5}{13}=\tfrac{2}{13}$. Multiplying by $P(A)$ uses the wrong conditioning event. …