A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and t

Q: Wait, why do we use the inequality $P(A \cup B) \le 1$ instead of an equality?

That is a fantastic question! In probability, the total probability of the sample space is exactly 1. The union $P(A \cup B)$ represents the probability that at least one event occurs. While it is possible for this union to be 1 (if the events cover the entire sample space), it can also be less than 1 if there are outcomes where neither A nor B is selected. Therefore, $P(A \cup B) \le 1$ is the correct constraint to establish an upper bound.

Q: Why is the intersection $P(A \cap B) \le 0.3$ used as a maximum value?

The problem states that the probability of both being selected is 'almost 0.3', which mathematically translates to $P(A \cap B) \le 0.3$. To find the maximum possible value for $P(B)$, we need to make the term we are subtracting, $P(A \cap B)$, as small as possible, or conversely, look at the boundary condition. By using the maximum value of the intersection, we are testing the most generous scenario for $P(B)$ to see if it can reach 0.9.

Q: If $P(B) = 0.9$ and $P(A) = 0.5$, why can't the intersection just be larger to make it work?

Great intuition, but remember that the intersection $P(A \cap B)$ is constrained by the individual probabilities. Specifically, $P(A \cap B)$ cannot be greater than $P(A)$ or $P(B)$. More importantly, the problem explicitly gives us the constraint $P(A \cap B) \le 0.3$. We must respect the constraints provided in the problem statement. If we were allowed to change the intersection, the answer might be different, but within the rules provided, 0.9 is simply out of reach.

Question

$A$ and $B$ are two candidates seeking admission in IIT. The probability that $A$ is selected is 0.5 and the probability that both $A$ and $B$ are selected is almost 0.3. Is it possible that the probability of $B$ getting selected is 0.9?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITian. Today, we stand at the intersection of logic and ambition. You are looking at a problem that seems simple, yet it tests the very foundation of how we perceive possibility.

We have two candidates, $A$ and $B$, each hoping to sail across the ocean of success. We are given the probability of $A$ being selected, $P(A) = 0.5$, and a constraint on the overlap, $P(A \cap B) \le 0.3$.

The question is: can $P(B)$ be $0.9$? Let's dive in.

### Visualizing the Sample Space

First, visualize the Venn diagram. The circle $A$ represents the event of $A$'s success, and circle $B$ represents $B$'s success.

The overlap, $P(A \cap B)$, is the region where both succeed. This is the heart of the problem; we are dealing with the geometry of chance.

### The Logic Bridge: The Addition Theorem

To connect these probabilities, we use a fundamental tool in probability theory: the Addition Theorem. This theorem states that the probability of the union of two events, $P(A \cup B)$, is equal to the sum of their individual probabilities minus the probability of their intersection.

Mathematically, we write this as:

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

This formula is elegant; it accounts for the fact that if we simply add the probabilities of $A$ and $B$, we count the overlap twice. We must subtract it once to get the true union.

### The Reality Check: The Axiom of Probability

Here is the crucial logic bridge. By the axioms of probability, the probability of any event—including the union of $A$ and $B$—can never exceed $1$.

So, $P(A \cup B) \le 1$. This is the ceiling of our reality.

If we substitute our addition formula into this inequality, we get:

$$P(A) + P(B) - P(A \cap B) \le 1$$

Now, let's plug in the known value of $P(A) = 0.5$. Our inequality becomes:

$$0.5 + P(B) - P(A \cap B) \le 1$$

### Isolating the Possibility

Our goal is to find the limits on $P(B)$. Let's isolate $P(B)$ step-by-step.

Subtracting $0.5$ from both sides, we get $P(B) - P(A \cap B) \le 0.5$. Adding $P(A \cap B)$ to both sides, we arrive at:

$$P(B) \le 0.5 + P(A \cap B)$$

Now, let's bring in the constraint on the intersection. We are given that $P(A \cap B) \le 0.3$.

To find the absolute maximum possible value for $P(B)$, we substitute this maximum limit of $0.3$ into our inequality:

$$P(B) \le 0.5 + 0.3$$

### The Final Verdict

Let's perform the final addition. $0.5 + 0.3 = 0.8$. This means the probability of candidate $B$ getting selected can never exceed $0.8$.

No matter how we arrange the events, $P(B)$ is capped at $0.8$. Now, let's answer the original question. Is it possible for the probability of $B$ getting selected to be $0.9$?

Since $0.9 > 0.8$, this violates the probability constraint. It is **absolutely impossible**.

If $P(B)$ were $0.9$, the total probability of their union would exceed $1$, which is a direct violation of probability theory. Keep this logic close to your heart, and you will navigate any problem with confidence.

Animated Solution for Mathematics - Probability: $A$ and $B$ are two candidates seeking admission in IIT. The probability that $A$ is selected is 0.5 and the probability that both $A$ and $B$ are selected is almost 0.3. Is it possible that the probability of $B$ getting selected is 0.9?

Visualized Solution

Visualizing the Sample Space

Identifying Given Values

The Addition Theorem of Probability

Applying the Probability Axiom

Substituting Known Values

Isolating $P (B)$

Applying the Intersection Constraint

Calculating the Upper Bound

The Final Verdict

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Visualizing the Sample Space

The Logic Bridge

The Reality Check

Isolating the Possibility

The Final Verdict

Similar Questions

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$A$ 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 the probability that exactly one of them occurs is $2/5$ and the probability that $A$ or $B$ occurs is $1/2$ , then the probability of both of them occur together is

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

Visualized Solution

Visualizing the Sample Space

Identifying Given Values

The Addition Theorem of Probability

Applying the Probability Axiom

Substituting Known Values

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(B)

Applying the Intersection Constraint

Calculating the Upper Bound

The Final Verdict

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Visualizing the Sample Space

The Logic Bridge

The Reality Check

Isolating the Possibility

The Final Verdict

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and B are two independent events. C is event in which exactly one of A or B occurs. Prove that P(C)≥P(A∪B)P(Aˉ∩Bˉ).

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Isolating $P (B)$

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

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

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