The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneous

Q: Wait, why is the sum of probabilities $P(\bar{A}) + P(\bar{B})$ greater than 1? Isn't probability always between 0 and 1?

That is a fantastic observation! Remember, the sum of probabilities of *mutually exclusive* events must be $\le 1$. However, $A$ and $B$ are not necessarily mutually exclusive. $P(\bar{A}) + P(\bar{B})$ is simply the sum of two individual probabilities. It can absolutely exceed 1, just as the sum of two numbers between 0 and 1 can exceed 1. You are not calculating a single probability here; you are calculating the sum of two separate probabilities.

Q: Can I use De Morgan's Laws here? Like, $P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B})$?

You absolutely can! De Morgan's Laws are powerful tools. However, the question asks for $P(\bar{A}) + P(\bar{B})$, not the probability of the union of the complements. Using De Morgan's would give you $P(\bar{A} \cup \bar{B}) = 1 - P(A \cap B) = 1 - 0.2 = 0.8$. But that is not what the question asks for. Stick to the direct approach using the complement rule for each event individually; it is much safer and more direct.

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

Think of it as a double-counting problem. When you add the area of circle $A$ and the area of circle $B$, the overlapping region (the intersection) is included in circle $A$ *and* it is included in circle $B$. By adding $P(A) + P(B)$, you have counted that middle overlap twice! To get the true area of the union (the total region covered by both), you must subtract that overlap once to correct the double-counting.

Question

The probability that at least one of the events $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.2$, then $P(\bar{A}) + P(\bar{B})$ is

Options:

(A) $0.4$

(B) $0.8$

(C) $1.2$

(D) $1.4$

(E) none

Sigma Percentile · Accepted Answer

### The Elegance of Probability: Unlocking the Union and the Complement

Welcome, fellow traveler on the JEE journey! Today, we are going to unravel a beautiful problem in probability. It might look like a simple algebraic exercise, but beneath the surface lies the fundamental geometry of sets and the logic of uncertainty.

### Phase 1: Visualizing the Landscape

Imagine a sample space—a vast, abstract canvas. On this canvas, we have two events, $A$ and $B$.

The problem gives us two vital pieces of information. First, the probability that at least one of these events occurs is $P(A \cup B) = 0.6$. In the language of sets, 'at least one' is the union—the entire region covered by both circles $A$ and $B$ combined.

Second, they occur simultaneously with a probability of $P(A \cap B) = 0.2$. This is the intersection, the precious, narrow overlap where both events happen at the exact same time.

### Phase 2: The Addition Theorem—The Great Corrector

Now, how do we connect these? We turn to the Addition Theorem of Probability:

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

Why do we subtract the intersection? Imagine you are painting the region $A$ and then painting the region $B$. When you paint $B$, you paint over the part that is already covered by $A$. You have counted that overlap twice!

To get the true area of the union, we must subtract the intersection once. Substituting our known values, we get:

$$0.6 = P(A) + P(B) - 0.2$$

With a simple algebraic shift, moving the $-0.2$ to the other side, we find the sum of the individual probabilities:

$$P(A) + P(B) = 0.6 + 0.2 = 0.8$$

Keep this value, $0.8$, close to your heart. It is the key to our final destination.

### Phase 3: The Complementary Leap

The question asks us to find $P(\bar{A}) + P(\bar{B})$. These are complementary events.

The complement rule is one of the most elegant tools in our arsenal: the probability of an event *not* happening is simply $1$ minus the probability of it happening. So, we have:

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

Let us substitute these into our target expression:

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

Now, let us group the terms. We have $1 + 1$, which is $2$, and then we have $-P(A) - P(B)$. If we factor out the negative sign, we get:

$$P(\bar{A}) + P(\bar{B}) = 2 - (P(A) + P(B))$$

### Phase 4: The Final Triumph

We are almost there! Remember that sum we calculated in Phase 2? We found that $P(A) + P(B) = 0.8$.

Let us plug that value into our expression:

$$P(\bar{A}) + P(\bar{B}) = 2 - 0.8$$

The subtraction is straightforward, yet satisfying:

$$P(\bar{A}) + P(\bar{B}) = 1.2$$

And there it is! The answer is **$1.2$**. It is a beautiful result, isn't it?

We navigated through the union, corrected for double-counting, utilized the complement rule, and arrived at the solution with logical precision. Never fear the complexity of a problem; break it down, visualize the geometry, and let the math guide you home.

Animated Solution for Mathematics - Probability: The probability that at least one of the events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.2$ , then $P (\overset{ˉ}{A}) + P (\overset{ˉ}{B})$ is

Select Answer:

Visualized Solution

Visualizing Events $A$ and $B$

The Union of Events

The Intersection of Events

The Addition Theorem of Probability

Substituting the Given Values

Isolating the Sum

Evaluating the Sum

Defining Complementary Events

Setting up the Target Expression

Rearranging the Terms

Final Substitution

Final Conclusion

The Sigma Insight: Addition and Multiplication Theorems

The Elegance of Probability

Phase 1

Phase 2

Phase 3

Phase 4

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