Let 0

Q: Wait, why does $P(A \cap B) = P(A)P(B)$ automatically mean independence?

Independence is defined by the condition that the occurrence of one event does not change the probability of the other. Mathematically, this is equivalent to $P(A \cap B) = P(A)P(B)$. If this holds, the events are independent. It is the fundamental definition, not just a coincidence!

Q: Can I always assume $P(A \cup B) = P(A) + P(B) - P(A)P(B)$?

No, absolutely not! That is only true when events are independent. The general formula is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. The problem *gives* you this specific condition to tell you that these events are independent. Always check the problem statement for such clues.

Q: Why is $P(A/B) = P(A)$?

By definition, $P(A/B) = P(A \cap B) / P(B)$. Since $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$. Substituting this gives $P(A)P(B) / P(B) = P(A)$. It makes sense: if $A$ and $B$ are independent, knowing $B$ happened gives you zero new information about $A$.

Question

Let $0 < P(A) < 1, 0 < P(B) < 1$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ then

Options:

(A) $P(B/A) = P(B) - P(A)$

(B) $P(A^c - B^c) = P(A^c) - P(B^c)$

(C) $P(A \cup B)^c = P(A^c)P(B^c)$

(D) $P(A/B) = P(A)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the world of mathematics. Today, we are not just solving a problem; we are uncovering a hidden symmetry in the fabric of probability.

Imagine you are standing before a Venn diagram, looking at two events, $A$ and $B$, within a sample space. We are given a specific, almost magical condition:

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

At first glance, this might look like just another algebraic expression, but to the trained eye, it is a whisper of a deeper truth.

### The Revelation of Independence

Let us recall our most trusted tool: the General Addition Theorem. It states that for any two events:

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

Now, look at the equation provided in the problem. The first two terms, $P(A)$ and $P(B)$, are identical to the general theorem. The only difference lies in the final term.

By comparing these two, we are forced to conclude that:

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

This is the golden key. This equality is the precise definition of **independent events**. It tells us that the occurrence of $A$ has absolutely no influence on the occurrence of $B$.

### Unraveling the Complement

Now that we have established that $A$ and $B$ are independent, let us consider the complement of the union, $P((A \cup B)^c)$. We know that the probability of any event's complement is $1$ minus the probability of the event itself:

$$P((A \cup B)^c) = 1 - P(A \cup B)$$

Substituting our given condition, we get:

$$P((A \cup B)^c) = 1 - [P(A) + P(B) - P(A)P(B)]$$

Distributing the negative sign gives us $1 - P(A) - P(B) + P(A)P(B)$. We can factorize this expression by grouping terms:

$$P((A \cup B)^c) = (1 - P(A)) - P(B)(1 - P(A))$$

Factoring out $(1 - P(A))$, we are left with:

$$P((A \cup B)^c) = (1 - P(A))(1 - P(B))$$

Since $1 - P(A) = P(A^c)$ and $1 - P(B) = P(B^c)$, we have proven that:

$$P((A \cup B)^c) = P(A^c)P(B^c)$$

### The Intuition of Conditional Probability

Finally, let us look at the conditional probability $P(A|B)$. The definition is:

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

Since we have already proven that $A$ and $B$ are independent, we know that $P(A \cap B) = P(A)P(B)$. Substituting this into our conditional probability formula:

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

As long as $P(B) > 0$, we can cancel $P(B)$ from the numerator and the denominator. What remains is simply:

$$P(A|B) = P(A)$$

This result is deeply intuitive. If $A$ and $B$ are independent, knowing that $B$ has occurred provides no new information about $A$. The probability of $A$ remains unchanged.

Animated Solution for Mathematics - Probability: Let $0 < P (A) < 1, 0 < P (B) < 1$ and $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ then

Select Answer:

Visualized Solution

Problem Setup

The Given Condition

General Addition Theorem

Comparing Equations

Independent Events

Checking Option (c)

Substituting the Union

Algebraic Expansion

Factorization

Option (c) is Correct

Checking Option (d)

Using Independence

Simplifying the Expression

Final Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Revelation of Independence

Unraveling the Complement

The Intuition of Conditional Probability

Similar Questions

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

For two given events $A$ and $B, P (A \cap B)$ is

$P (A \cup B) = P (A \cap B)$ if and only if the relation between $P (A)$ and $P (B)$ is .........

If $P (B) = \frac{3}{4}, P (A \cap B \cap \overset{ˉ}{C}) = \frac{1}{3}$ and $P (\overset{ˉ}{A} \cap B \cap \overset{ˉ}{C}) = \frac{1}{3}$ , then $P (B \cap C)$ is

$A$ and $B$ are events such that $P (A \cup B) = 3/4, P (A \cap B) = 1/4, P (\overset{ˉ}{A}) = 2/3$ then $P (\overset{ˉ}{A} \cap B)$ is

If $M$ and $N$ are any two events, the probability that exactly one of them occurs is

$A$ and $B$ are two independent events. $C$ is event in which exactly one of $A$ or $B$ occurs. Prove that $P (C) \geq P (A \cup B) P (\overset{ˉ}{A} \cap \overset{ˉ}{B})$ .

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 $P (B) = .........$

Animated Solution for Mathematics - Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 0<P(A)<1,0<P(B)<1 and P(A∪B)=P(A)+P(B)−P(A)P(B) then