For any two events A and B in a sample space
(a) P(A/B) ≥ (P(A) + P(B) - 1)/P(B), P(B) ≠ 0 is always

Q: Why can't we just assume $P(A \cap B) = P(A)P(B)$ for all cases?

That is a classic trap! The product rule $P(A \cap B) = P(A)P(B)$ is only valid if the events are independent. In a general JEE problem, you must never assume independence unless it is explicitly stated or implied by the context. Always rely on the general Addition Theorem first.

Q: Why is $P(A \cap \bar{B}) = P(A) - P(A \cap B)$ always true?

Think of it as set subtraction. The set $A$ is composed of two disjoint parts: the part that is also in $B$ (which is $A \cap B$) and the part that is not in $B$ (which is $A \cap \bar{B}$). Thus, $A = (A \cap B) \cup (A \cap \bar{B})$. Since these are disjoint, their probabilities add up, leading to $P(A) = P(A \cap B) + P(A \cap \bar{B})$.

Q: Why does the formula $P(A \cup B) = 1 - P(\bar{A})P(\bar{B})$ fail for disjoint events?

That formula is derived specifically for independent events. For disjoint events, $P(A \cap B) = 0$, which leads to $P(A \cup B) = P(A) + P(B)$. The formula $1 - P(\bar{A})P(\bar{B})$ simplifies to $P(A) + P(B) - P(A)P(B)$. These two expressions are only equal if $P(A)P(B) = 0$, which is not guaranteed for disjoint events.

Question

For any two events $A$ and $B$ in a sample space

Options:

(A) $P(A/B) \ge \frac{P(A) + P(B) - 1}{P(B)}, P(B) 
eq 0$ is always true

(B) $P(A \cap \bar{B}) = P(A) - P(A \cap B)$ does not hold

(C) $P(A \cup B) = 1 - P(\bar{A})P(\bar{B})$, if $A$ and $B$ are independent

(D) $P(A \cup B) = 1 - P(\bar{A})P(\bar{B})$, if $A$ and $B$ are disjoint.

Sigma Percentile · Accepted Answer

### Analyzing the Conditional Probability Bound

Let us begin with Option (a). We are asked about the conditional probability $P(A|B)$.

By definition, this is the ratio of the intersection to the probability of the condition:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

To find a lower bound for this, we need to understand the minimum possible size of the intersection $A \cap B$. We turn to the Addition Theorem:

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

The union of two events $A \cup B$ is a subset of the sample space $S$, so its probability can never exceed $1$. Thus, we have the inequality:

$$P(A) + P(B) - P(A \cap B) \le 1$$

Rearranging this, we find the fundamental inequality:

$$P(A \cap B) \ge P(A) + P(B) - 1$$

When we substitute this back into our conditional probability formula, we obtain:

$$P(A|B) \ge \frac{P(A) + P(B) - 1}{P(B)}$$

This confirms that Option (a) is a **universal truth**.

### The Set Difference Identity

Next, let us examine Option (b). It claims that $P(A \cap \bar{B}) = P(A) - P(A \cap B)$ does not hold.

$A \cap \bar{B}$ represents the region where $A$ occurs but $B$ does not—the "Only $A$" region. If you visualize a Venn diagram, the entire circle $A$ is composed of two disjoint pieces: the intersection $A \cap B$ and the "Only $A$" region $A \cap \bar{B}$.

Therefore, the following identity is **rock-solid**:

$$P(A) = P(A \cap B) + P(A \cap \bar{B})$$

Since the statement in Option (b) claims this identity does not hold, the statement itself is **incorrect**.

### The Independence Trap

Now, we arrive at Option (c). We are given that $A$ and $B$ are independent, which is a powerful condition.

If $A$ and $B$ are independent, then their complements $\bar{A}$ and $\bar{B}$ are also independent. This implies:

$$P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B})$$

By De Morgan's Law, we know that $\overline{A \cup B} = \bar{A} \cap \bar{B}$. Since the probability of an event is $1$ minus the probability of its complement, we have:

$$P(A \cup B) = 1 - P(\overline{A \cup B}) = 1 - P(\bar{A} \cap \bar{B})$$

Substituting our independence result, we get:

$$P(A \cup B) = 1 - P(\bar{A})P(\bar{B})$$

This confirms that Option (c) is **correct**.

### The Disjoint Delusion

Finally, Option (d) suggests the same formula works for disjoint events. Disjoint events imply that $P(A \cap B) = 0$.

In this case, the addition rule simplifies to:

$$P(A \cup B) = P(A) + P(B)$$

However, our formula from Option (c) expands to:

$$P(A \cup B) = P(A) + P(B) - P(A)P(B)$$

For these to be equal, we would require $P(A)P(B) = 0$. Since disjoint events do not necessitate this condition, Option (d) is **false**.

Remember, independence and disjointness are two very different worlds. Never confuse them.

Animated Solution for Mathematics - Probability: For any two events $A$ and $B$ in a sample space

Select Answer:

Visualized Solution

Visualizing the Sample Space

Option (a): Conditional Probability

The Addition Theorem

Bounding the Intersection

Concluding Option (a)

Option (b): Set Difference

Evaluating Option (b)

Option (c): Independent Events

De Morgan's Law for Option (c)

Option (d): Disjoint Events

Evaluating Option (d)

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Conditional Probability Bound

The Set Difference Identity

The Independence Trap

The Disjoint Delusion

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Animated Solution for Mathematics - Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For any two events A and B in a sample space