Let A and B be two events such that P(A) = 0.3 and P(A ∪ B) = 0.8. If A and B are independent events

Q: Wait, why do we subtract $P(A \cap B)$ in the addition theorem? Isn't it just $P(A) + P(B)$?

That is a classic trap! Think of it like this: if you have two overlapping circles in a Venn diagram, the region where they overlap is part of circle A and also part of circle B. If you add the entire area of A and the entire area of B, you have counted that overlapping region twice. We subtract the intersection $P(A \cap B)$ exactly once to correct this double-counting.

Q: How do I know for sure that $P(A \cap B) = P(A) \cdot P(B)$?

This is the formal definition of independent events. Independence means that the occurrence of event A provides absolutely no information about the likelihood of event B occurring. Mathematically, this 'lack of influence' is captured by the multiplication rule. If they were not independent, we would need conditional probability, but here, the independence is your strongest tool!

Q: Is there any other way to solve this besides algebra?

You could visualize this with a Venn diagram, but algebra is much more precise and less prone to drawing errors. By setting up the equation $0.8 = 0.3 + P(B) - 0.3 \cdot P(B)$, you are essentially performing a systematic search for the missing piece of the puzzle. Stick to the algebra—it is the most reliable path to the correct answer.

Question

Let $A$ and $B$ be two events such that $P(A) = 0.3$ and $P(A \cup B) = 0.8$. If $A$ and $B$ are independent events then $P(B) = .........$

Sigma Percentile · Accepted Answer

### The Elegant Dance of Independent Events

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a probability problem; we are uncovering the hidden architecture of chance.

Probability is often seen as a collection of disjointed formulas, but when you look closer, it is a beautiful, logical structure. Let us break down this problem, step by step, and see how the pieces fit together.

### Phase 1: The Foundation of Inclusion-Exclusion

We start with the Addition Theorem of Probability. Imagine you are looking at a Venn diagram with two events, $A$ and $B$.

You want to find the probability of their union, $P(A \cup B)$, which represents the probability that *at least one* of these events occurs. The formula is intuitive: 
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Why the subtraction? It is all about bookkeeping. When you add the probability of $A$ to the probability of $B$, you are counting the region where they overlap—the intersection $P(A \cap B)$—twice.

To get the true area of the union, we must remove that extra count. This is the fundamental bookkeeping rule of probability.

### Phase 2: The Power of Independence

Now, the problem gives us a gift: $A$ and $B$ are independent. In the world of probability, independence is a superpower.

It means that the outcome of $A$ has zero impact on the outcome of $B$. Mathematically, this transforms our intersection term:
$$P(A \cap B) = P(A) \cdot P(B)$$

This is the bridge we need. By substituting this into our addition theorem, we turn a problem with two unknowns into a problem with only one: 
$$P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$$

### Phase 3: The Algebraic Resolution

Now, let us bring in the numbers. We know $P(A) = 0.3$ and $P(A \cup B) = 0.8$. Our equation becomes:

$$0.8 = 0.3 + P(B) - 0.3 \cdot P(B)$$

This is where many students rush and stumble. Take a breath. We want to isolate $P(B)$. First, move the $0.3$ to the left side:

$$0.8 - 0.3 = P(B) - 0.3 \cdot P(B)$$

$$0.5 = P(B) \cdot (1 - 0.3)$$

See the elegance? By factoring out $P(B)$, we simplify the right side to $P(B) \cdot 0.7$. Now, the final step is just simple division:

$$P(B) = \frac{0.5}{0.7} = \frac{5}{7}$$

### The Takeaway

Look at what we have achieved. We started with a set of abstract conditions and, by applying the logic of independence and the addition theorem, we arrived at a clean, rational fraction.

This is the essence of JEE mathematics: taking complex, seemingly overwhelming scenarios and reducing them to their core, elegant truths. Keep practicing this mindset, and you will find that no problem is too daunting.

**The final answer is $P(B) = \frac{5}{7}$.**

Animated Solution for Mathematics - Probability: Let $A$ and $B$ be two events such that $P (A) = 0.3$ and $P (A \cup B) = 0.8$ . If $A$ and $B$ are independent events then $P (B) = .........$

Visualized Solution

Given Information

Addition Theorem of Probability

Condition for Independent Events

Modifying the Formula

Substituting Given Values

Rearranging the Equation

Factoring out $P (B)$

Simplifying the Expression

Final Answer

The Sigma Insight: Addition and Multiplication Theorems

The Elegant Dance of Independent Events

Phase 1

Phase 2

Phase 3

The Takeaway

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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; } Let A and B be two events such that P(A)=0.3 and P(A∪B)=0.8. If A and B are independent events then P(B)=.........