Problem 127 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(A\) and \(B\) are two events such that \(P(A)=\frac{1}{2}\) and \(P(B)=\frac{2}{3}\), then (A) \(P(A \cup B) \geq \frac{2}{3}\) (B) \(P\left(A \cap B^{\prime}\right) \leq \frac{1}{3}\) (C) \(1 / 6 \leq P(A \cap B) \leq \frac{1}{2}\) (D) \(1 / 6 \leq P\left(A^{\prime} \cap B\right) \leq \frac{1}{2}\)

Step-by-Step Solution

Answer

All options A, B, C, and D are correct.

Step 1: Identify Given Information

We are given two events, \(A\) and \(B\), with probabilities:\[ P(A) = \frac{1}{2} \] \[ P(B) = \frac{2}{3} \] We need to evaluate statements about their union and intersection.

Step 2: Calculate Maximum Value of \( P(A \cap B) \)

Using the formula for the union of two probabilities: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] we find the maximum value of \(P(A \cap B)\) by setting \( P(A \cup B) = 1 \) (the highest possible value). Thus, \[ \frac{1}{2} + \frac{2}{3} - P(A \cap B) = 1 \] Which gives \[ P(A \cap B) = \frac{1}{6} \] as the minimum possible value.

Step 3: Calculate Minimum Value of \( P(A \cap B) \)

The minimum value of \(P(A \cap B)\) occurs when \(P(A \cup B)\) is minimized. Using the inequality \(P(A \cup B) \geq max(P(A), P(B))\), we find \(P(A \cup B)\) can be as low as \( \frac{2}{3}\). Solving for \(P(A \cap B)\) in \( \frac{1}{2} + \frac{2}{3} - P(A \cap B) \geq \frac{2}{3} \), we find \(P(A \cap B) \leq \frac{1}{2}\). Concluding \( \frac{1}{6} \leq P(A \cap B) \leq \frac{1}{2} \).

Step 4: Check Validity of Option A and B

For option (A): Since \( P(A \cup B) \geq \max(P(A), P(B)) = \frac{2}{3} \), statement (A) is correct. For option (B): \( P(A \cap B') = P(A) - P(A \cap B) \). With \( \frac{1}{6} \leq P(A \cap B) \leq \frac{1}{2} \), \( P(A \cap B') \leq \frac{1}{3} \) is correct.

Step 5: Evaluate Option C

For statement (C): We have already established \( \frac{1}{6} \leq P(A \cap B) \leq \frac{1}{2} \), so (C) is correct.

Step 6: Evaluate Option D

For statement (D): \( P(A' \cap B) = P(B) - P(A \cap B) \). With \(P(A \cap B) \geq \frac{1}{6}\), this gives \(P(A' \cap B) \leq \frac{2}{3} - \frac{1}{6} = \frac{1}{2}\). Since \(P(A' \cap B) \) starts at \( \frac{1}{6}\), statement (D) is correct.

Key Concepts

Union of Events

The union of events in probability theory is a fundamental concept. To understand it, imagine two events, event \(A\) and event \(B\). The union, represented as \(A \cup B\), encompasses all the possible outcomes that belong to either \(A\), \(B\), or both. It's like bundling up everything that could happen in either event.

The formula used to calculate this is:

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

This formula combines the probabilities of events \(A\) and \(B\), but helps avoid double-counting the overlap, \(P(A \cap B)\), which is the intersection discussed next.

In simple words, if you want to know how likely it is for either of these events to happen, you'd consider their union probability.

Intersection of Events

The intersection of events is when both event \(A\) and event \(B\) occur at the same time. This is denoted as \(A \cap B\). Think of it as the shared space in a Venn diagram where both circles overlap. In probability, the intersection is crucial because it directly speaks about the likelihood of both events happening together.

The probability of the intersection, \(P(A \cap B)\), can be calculated if we have certain information about these events:

• The formula is: \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\)

By understanding this, we grasp not only the overlap of events but also know how events are interrelated.

Contrary to just having event \(A\) or event \(B\) happen, intersections tell us about the precise chance of both happening—a critical insight for deeper probability analysis.

Complementary Events

Complementary events refer to situations where an event does not occur. If you're interested in an event \(A\), its complement, denoted as \(A'\), consists of all outcomes that are not in \(A\). It's like flipping a switch—if \(A\) is the light being on, \(A'\) is the light being off.

The principle behind complements is:

• \(P(A') = 1 - P(A)\)

This equation arises because the sum of the probabilities of an event and its complement is always equal to 1. This makes sense intuitively, because if \(A\) happens or \(A'\) happens, there are no other outcomes.

In the context of our problem, using complements provides alternative avenues to approach a problem, as it considers everything not covered by the original event. This concept is invaluable in analyzing uncertainty and probability extensively.