Problem 71 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The probabilites of three events \(A, B\) and \(C\) are \(P(A)=0.6, P(B)=0.4\) and \(P(C)=0.5 .\) If \(P(A \cup B)\) \(=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)=0.2\) and \(P(A \cup\) \(B \cup C) \geq 0.85\), then (A) \(0.2 \leq P(B \cap C) \leq 0.35\) (B) \(0.5 \leq P(B \cap C) \leq 0.85\) (C) \(0.1 \leq P(B \cap C) \leq 0.35\) (D) none of these

Step-by-Step Solution

Answer

(C) is the closest correct answer, but the problem contains contradictions.

Step 1: Write Down Given Probabilities

We have the following probabilities given: \(P(A) = 0.6\), \(P(B) = 0.4\), \(P(C) = 0.5\), \(P(A \cup B) = 0.8\), \(P(A \cap C) = 0.3\), \(P(A \cap B \cap C) = 0.2\), and \(P(A \cup B \cup C) \geq 0.85\).

Step 2: Use Inclusion-Exclusion Principle

To find \(P(B \cap C)\), we will use the inclusion-exclusion principle for three sets: \[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - \]\[ P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] Given \(P(A \cup B \cup C) \geq 0.85\), proceed to the next step.

Step 3: Express Known Intersections

From the problem, \(P(A \cap B\) can be found using: \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] Plugging in the values: \[0.8 = 0.6 + 0.4 - P(A \cap B)\] Solve for \(P(A \cap B)\): \(P(A \cap B) = 0.2\)

Step 4: Solve for \(P(B \cap C)\)

Insert known values into the inclusion-exclusion principle:\[0.85 \leq 0.6 + 0.4 + 0.5 - 0.2 - 0.3 - P(B \cap C) + 0.2\] Simplify and solve for \(P(B \cap C)\): \[0.85 \leq 1.2 - 0.3 - P(B \cap C)\] \[0.85 \leq 0.9 - P(B \cap C)\]\[P(B \cap C) \leq 0.05\] which indicates a contradiction since \(P(B \cap C)\geq P(A \cap B \cap C) = 0.2\).

Step 5: Check for Possible Options

Calculate possible values for \(P(B \cap C)\) within the constraints given: Since the calculations yield \(P(B \cap C)\) should be at least 0.2 but cannot be consistent with \(0.05\) from step 4, notice this is a contradiction with option (D), meaning given options do not satisfy logical condition proven during calculations. Option (C) could be closest to correct based on problem setup, recognizing potential error on bounds.

Key Concepts

Inclusion-Exclusion Principle

The Inclusion-Exclusion Principle is a fundamental concept in probability theory that helps calculate the probability of the union of multiple events. When dealing with probabilities of events that have overlaps, such as events A, B, and C, there might be areas that are counted more than once if we simply add up the individual probabilities. The Inclusion-Exclusion Principle corrects this by removing the probabilities of the intersections that have been counted too many times and adding back the intersections that were subtracted too much.

In formal terms, for three events A, B, and C, the principle states:

• The probability of at least one of the events occurring is: \[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\]

This formula ensures that every intersection between events is considered properly, preventing overcounting or undercounting. In our original problem, this principle was pivotal in identifying a range of possible values for \( P(B \cap C) \) and in highlighting a contradiction, leading to insights about correct probabilities for event intersections.

Intersection of Events

The intersection of events is an important concept in probability that describes the event in which two or more specific events occur simultaneously. For instance, the intersection of events A and B, denoted as \( A \cap B \), represents the outcome where both events occur together.

In probability terms, the intersection of two events can often be calculated by knowing the probability of each event and any known probabilities related to their union or other intersections. In mathematical notation, for example, the probability of intersection for two events can be expressed as:

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

This formula subtracts the probability of the union from the sum of individual probabilities, accounting for overlap. This concept proved useful in simplifying calculations, as demonstrated in the original solution when finding \( P(A \cap B) \). Additionally, knowing probabilities of intersections like \( P(A \cap B \cap C) \) provides constraints on possible values of other intersections, like \( P(B \cap C) \), as outlined in the given exercise.

Union of Events

The union of events in probability represents the scenario where at least one of the considered events occurs. It is denoted as \( A \cup B \cup C \) for multiple events A, B, and C, meaning that event A, event B, event C, or any combination of them happens.

Calculating the probability of the union is essential in understanding the likelihood of combinations of events happening. The key formula often used is:

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

This formula accounts for the overlap or intersection, ensuring that the instances in which both A and B happen are not counted twice.
In the context of the original exercise, knowing \( P(A \cup B) = 0.8 \) played a crucial role in uncovering the probability of their intersection. Furthermore, using the probabilities of each event and their known intersections helped determine other intersection probabilities needed for making sense of the problem's constraints.