Problem 39 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A, B, C\), be pairwise independent events with \(P(\mathrm{C})>0\) and \(P(A \cap B \cap C)=0\). Then \(P\left(A^{c} \cap B^{c} / C\right)\). [2011RS] (a) \(\mathrm{P}\left(B^{c}\right)-\mathrm{P}(B)\) (b) \(P\left(A^{c}\right)+P\left(B^{c}\right)\) (c) \(P\left(A^{c}\right)-P\left(B^{c}\right)\) (d) \(P\left(A^{c}\right)-P(B)\)

Step-by-Step Solution

Answer

Choice (d): \(P(A^{c}) - P(B)\).

Step 1: Calculate Conditional Probability

The problem asks for \(P\left(A^{c} \cap B^{c} \mid C\right)\), which is defined as the conditional probability \[ P\left(A^{c} \cap B^{c} \mid C\right) = \frac{P\left(A^{c} \cap B^{c} \cap C\right)}{P(C)}. \]

Step 2: Use Given Conditional

The problem states that \(P(A \cap B \cap C) = 0\). It follows that \[ P\left(A^{c} \cap B^{c} \cap C\right) = P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) = P(C) - P(A \cap C) - P(B \cap C). \]

Step 3: Pairwise Independence Application

Since events \(A, B, C\) are pairwise independent, \(P(A \cap C) = P(A)P(C)\) and \(P(B \cap C) = P(B)P(C)\). Thus, \[ P\left(A^{c} \cap B^{c} \cap C\right) = P(C) - P(A)P(C) - P(B) P(C). \]

Step 4: Normalize to Find Conditional Probability

Using \(P\left(A^{c} \cap B^{c} \mid C\right) = \frac{P\left(A^{c} \cap B^{c} \cap C\right)}{P(C)}\), substitute the result from Step 3: \[ P\left(A^{c} \cap B^{c} \mid C\right) = 1 - P(A) - P(B). \]

Step 5: Conclude the Answer

From the calculations, we can conclude that \[ P\left(A^{c} \cap B^{c} \mid C\right) = P(A^{c}) - P(B). \] The correct choice aligns with answer choice (d).

Key Concepts

Pairwise Independence

Pairwise independence is an important concept in probability theory. It refers to a situation where two events, say \(A\) and \(B\), do not influence each other in terms of occurrence. This means the probability of both events happening together is simply the product of their individual probabilities. It is mathematically expressed as \(P(A \cap B) = P(A) \cdot P(B)\).
When three events \(A, B,\) and \(C\) are pairwise independent, any pair among them \((A, B), (B, C), (A, C)\) satisfy the condition \(P(A \cap B) = P(A) \cdot P(B)\).
It is important to note that pairwise independence is weaker than full independence, which would require all three events' joint probability to equal the product of all individual probabilities. Full independence implies pairwise independence, but not the other way around. This distinction often plays a crucial role in solving probability problems, like determining conditional probabilities, because it allows for simplifying assumptions without requiring full independence.

Complementary Events

Complementary events are a basic concept in probability that helps simplify understanding. If you have an event \(A\), the complementary event, noted as \(A^{c}\), includes all the possible outcomes that are not in \(A\). In other words, \(A^{c}\) represents the event that \(A\) does not happen.
The relationship between an event and its complement can be summarized with \(P(A) + P(A^{c}) = 1\), indicating that the total probability must equal 1.
This concept is particularly useful in problems requiring conditional probabilities, like finding \(P(A^{c} \cap B^{c} \mid C)\). Knowing the properties of complementary events allows us to approach a problem with the understanding that the sum of probabilities of mutually exclusive events—like an event and its complement—must tally to one. Simplifying probabilities using their complements can often provide a clearer path to a solution, as is seen in cases involving conditional probability computations.

Intersection of Events

The intersection of events is a key concept in probability, used to describe the occurrence of two or more events at the same time. For two events \(A\) and \(B\), their intersection, denoted by \(A \cap B\), is the event that both \(A\) and \(B\) occur simultaneously. In the language of probability, this is represented by \(P(A \cap B)\).
Understanding intersections is essential when dealing with conditional probabilities, as these often involve calculating the likelihood of one event given the occurrence of another. The formula \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) shows how conditional probability depends largely on the intersection of the events.
In the given problem, calculating \(P(A^{c} \cap B^{c} \mid C)\) involved assessing the intersection of the complements of \(A\) and \(B\) with another event \(C\). Recognizing and manipulating intersected events is crucial to obtaining accurate probability calculations, particularly when dealing with conditional scenarios where one event alters the probability of another.