Let A and B be two events such that P(A ∪ B) = 5/6, P(A ∩ B) = 1/4 and P(Ā) = 1/4, where Ā stands fo

Q: Wait, why do we subtract the intersection $P(A \cap B)$ in the Addition Theorem?

Think of the Venn diagram. When you add the area of circle $A$ and the area of circle $B$, you are counting the overlapping region (the intersection) twice—once as part of $A$ and once as part of $B$. Subtracting $P(A \cap B)$ once corrects this double-counting, giving you the true area of the union.

Q: What does it physically mean for two events to be 'independent'?

Independence means that the occurrence of event $A$ provides absolutely no information about whether event $B$ will occur. Mathematically, this is captured by the condition $P(A \cap B) = P(A) \cdot P(B)$. If this holds, the events are independent.

Q: Can two events be both mutually exclusive and independent?

Generally, no. If two events are mutually exclusive, $P(A \cap B) = 0$. If they are also independent, then $P(A) \cdot P(B) = 0$. This implies that at least one of the events must have a probability of zero. So, unless one of the events is impossible, they cannot be both mutually exclusive and independent.

Question

Let $A$ and $B$ be two events such that $P(A \cup B) = 5/6, P(A \cap B) = 1/4$ and $P(\bar{A}) = 1/4$, where $\bar{A}$ stands for the complement of the event $A$. Then the events $A$ and $B$ are

Options:

(A) independent but not equally likely

(B) independent and equally likely

(C) mutually exclusive and independent

(D) equally likely but not independent

Sigma Percentile · Accepted Answer

### The Architecture of Uncertainty

Welcome, fellow explorer of the mathematical universe! Today, we are not just solving a probability problem; we are peeling back the layers of uncertainty to reveal the hidden structure of events.

Probability is the language of the unknown. In this problem, we are given a set of clues about two events, $A$ and $B$, living within a sample space $S$. Our mission is to deduce their nature: are they independent? Are they equally likely?

### Phase 1: Unlocking the Hidden Variables

We start with the raw data provided: $P(A \cup B) = \frac{5}{6}$, $P(A \cap B) = \frac{1}{4}$, and $P(\bar{A}) = \frac{1}{4}$.

Before we can analyze the relationship between $A$ and $B$, we must know their individual probabilities. We are given $P(\bar{A})$, the probability of the complement of $A$.

The complement rule is one of the most elegant symmetries in probability: an event either happens or it does not. Thus, the sum of their probabilities must be unity. We write this as:

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

Substituting our known value, we find:

$$P(A) = 1 - \frac{1}{4} = \frac{3}{4}$$

With $P(A)$ in our toolkit, we turn to the Addition Theorem of Probability to find $P(B)$. This theorem is the bridge between the union, the individual events, and their intersection:

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

We have all the pieces of this puzzle except $P(B)$. Let us substitute the values we have:

$$\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}$$

Simplifying the right side, we combine the fractions $\frac{3}{4} - \frac{1}{4}$, which yields $\frac{2}{4}$, or $\frac{1}{2}$. Our equation now stands as:

$$\frac{5}{6} = \frac{1}{2} + P(B)$$

Solving for $P(B)$ is now a simple matter of subtraction:

$$P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}$$

### Phase 2: The Test of Independence

Now that we have established $P(A) = \frac{3}{4}$ and $P(B) = \frac{1}{3}$, we can test the nature of these events. The JEE loves to test your conceptual clarity on independence.

Two events are independent if, and only if, the probability of their intersection is the product of their individual probabilities:

$$P(A \cap B) = P(A) \cdot P(B)$$

Let us calculate the product $P(A) \cdot P(B)$:

$$P(A) \cdot P(B) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$$

Look at that! The result is exactly $\frac{1}{4}$, which matches the given $P(A \cap B)$. The condition is satisfied. We have mathematically proven that events $A$ and $B$ are **independent**.

### Phase 3: The Verdict

Finally, we must determine if the events are 'equally likely.' This is a straightforward comparison. Two events are equally likely if their probabilities are identical, i.e., $P(A) = P(B)$.

Comparing our results:

$$P(A) = \frac{3}{4} \approx 0.75$$

$$P(B) = \frac{1}{3} \approx 0.33$$

Since $\frac{3}{4} 
eq \frac{1}{3}$, the events are clearly **not equally likely**.

### Conclusion

Through this methodical journey, we have uncovered the truth: the events $A$ and $B$ are independent, but they are not equally likely. This problem serves as a beautiful reminder that in probability, as in life, we must look past the surface to understand the underlying relationships.

Animated Solution for Mathematics - Probability: Let $A$ and $B$ be two events such that $P (A \cup B) = 5/6, P (A \cap B) = 1/4$ and $P (\overset{ˉ}{A}) = 1/4$ , where $\overset{ˉ}{A}$ stands for the complement of the event $A$ . Then the events $A$ and $B$ are

Select Answer:

Visualized Solution

Visualizing the Events

Finding $P (A)$

Calculating $P (A)$

The Addition Theorem

Substituting Known Values

Simplifying the Equation

Solving for $P (B)$

Checking for Independence

Calculating $P (A) \cdot P (B)$

Conclusion on Independence

Checking for Equally Likely Events

Conclusion on Equally Likely

Final Answer

The Sigma Insight: Addition and Multiplication Theorems

The Architecture of Uncertainty

Phase 1

Phase 2

Phase 3

Conclusion

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