If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and P(A ∩ B̅) = 0.5, where B̅ denotes the

Q: Wait, why is $P(A \cap \bar{B})$ the 'A only' region? Shouldn't it be $P(A) - P(B)$?

That is a common trap! $P(A) - P(B)$ only works if $B$ is a subset of $A$. In general, $P(A) - P(B)$ doesn't account for the intersection correctly. The region $A \cap \bar{B}$ is defined as 'elements in $A$ AND elements NOT in $B$'. Since the intersection $A \cap B$ is part of $A$, removing $B$ (which includes the intersection) leaves you exactly with the part of $A$ that does not overlap with $B$. Hence, $P(A) = P(A \cap \bar{B}) + P(A \cap B)$ is the correct geometric decomposition.

Q: Why does $B \cap \bar{B}$ become an empty set? Is that always true?

Yes, it is always true! By definition, $\bar{B}$ is the complement of $B$, meaning it contains everything in the sample space that is NOT in $B$. Therefore, it is impossible for an element to be in both $B$ and $\bar{B}$ simultaneously. Their intersection is the empty set, $\emptyset$, and its probability is $0$. This is a powerful tool for simplifying complex set expressions!

Q: I'm confused about the denominator $P(A \cup \bar{B})$. Can I just add $P(A)$ and $P(\bar{B})$?

Not quite! Remember the general addition theorem: $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$. If you just add $P(A)$ and $P(\bar{B})$, you are double-counting the region where $A$ and $\bar{B}$ overlap. You must subtract the intersection $P(A \cap \bar{B})$ to get the correct probability.

Question

If $A$ and $B$ are two events such that $P(A) = 0.7$, $P(B) = 0.4$ and $P(A \cap \bar{B}) = 0.5$, where $\bar{B}$ denotes the complement of $B$, then $P(B | (A \cup \bar{B}))$ is equal:-

Options:

(A) $\frac{1}{4}$

(B) $\frac{1}{2}$

(C) $\frac{1}{6}$

(D) $\frac{1}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Chance: A Journey Through Probability

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a probability problem; we are mapping a territory.

Probability is often seen as a dry collection of formulas, but at its heart, it is the art of visualizing relationships. When we look at events $A$ and $B$, we are looking at two distinct regions in a universal set. Let us embark on this journey to find $P(B | (A \cup \bar{B}))$.

### Phase 1: Decoding the Map

Imagine a universal set $U$. Inside, we have two circles, $A$ and $B$, overlapping like a Venn diagram. We are given $P(A) = 0.7$ and $P(B) = 0.4$.

The most critical piece of information is $P(A \cap \bar{B}) = 0.5$. Geometrically, $\bar{B}$ is everything outside circle $B$. So, $A \cap \bar{B}$ is the portion of $A$ that refuses to touch $B$, often called the 'A only' region.

By understanding this, we see that the total probability of $A$ is the sum of the 'A only' region and the intersection $A \cap B$. Mathematically, we write:

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

Substituting our known values, we get $0.7 = 0.5 + P(A \cap B)$. A quick subtraction reveals that $P(A \cap B) = 0.2$. We have successfully identified the heart of the overlap!

### Phase 2: The Conditional Challenge

Now, we face the target: $P(B | (A \cup \bar{B}))$. The definition of conditional probability is our compass here:

$$P(X | Y) = \frac{P(X \cap Y)}{P(Y)}$$

Here, our $X$ is $B$, and our $Y$ is $(A \cup \bar{B})$. The numerator is $P(B \cap (A \cup \bar{B}))$.

Using the distributive law of sets, we distribute the intersection over the union:

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

Think about $B \cap \bar{B}$. Can an event happen and not happen at the same time? It is the empty set, $\emptyset$. Thus, our numerator simplifies beautifully to $P(A \cap B)$, which we already know is $0.2$.

### Phase 3: The Final Calculation

We have the numerator; now for the denominator: $P(A \cup \bar{B})$. Using the addition theorem, we have:

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

We know $P(A) = 0.7$ and $P(A \cap \bar{B}) = 0.5$. Since $P(B) = 0.4$, the complement rule tells us $P(\bar{B}) = 1 - 0.4 = 0.6$.

Plugging these into our equation:

$$P(A \cup \bar{B}) = 0.7 + 0.6 - 0.5 = 0.8$$

Finally, we bring it all together:

$$P(B | (A \cup \bar{B})) = \frac{0.2}{0.8} = \frac{1}{4}$$

And there it is! Through careful visualization and the elegant application of set theory, we have arrived at the solution. The final answer is **0.25** or **1/4**.

Animated Solution for Mathematics - Probability: If $A$ and $B$ are two events such that $P (A) = 0.7$ , $P (B) = 0.4$ and $P (A \cap \overset{ˉ}{B}) = 0.5$ , where $\overset{ˉ}{B}$ denotes the complement of $B$ , then $P (B ∣ (A \cup \overset{ˉ}{B}))$ is equal:-

Select Answer:

Visualized Solution

Visualizing the Events

Decoding $P (A \cap \overset{ˉ}{B})$

Finding the Intersection

Substituting Values

Calculating $P (A \cap B)$

Target Conditional Probability

Applying the Formula

Simplifying the Numerator

Evaluating the Numerator

Analyzing the Denominator

Finding $P (\overset{ˉ}{B})$

Calculating the Denominator

Evaluating the Denominator

Final Probability

The Sigma Insight: Conditional Probability

The Geometry of Chance

Phase 1

Phase 2

Phase 3

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