Problem 24 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) be three events, which are pair- wise independence and \(\bar{E}\) denotes the complement of an event E. If \(\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0\) and \(\mathrm{P}(\mathrm{C})>0\), then \(\mathrm{P}[(\bar{A} \cap \bar{B}) \mid \mathrm{C}]\) is equal to. \(\quad\) Online April 16, 2018] (a) \(\mathrm{P}(\mathrm{A})+\mathrm{P}(\bar{B})\) (b) \(\mathrm{P}(\bar{A})-\mathrm{P}(\bar{B})\) (c) \(\mathrm{P}(\bar{A})-\mathrm{P}(\mathrm{B})\) (d) \(\mathrm{P}(\bar{A})+\mathrm{P}(\bar{B})\)

Step-by-Step Solution

Answer

The answer is (d): \(P(\bar{A}) + P(\bar{B})\).

Step 1: Understand Pairwise Independence

The events \(A\), \(B\), and \(C\) are pairwise independent. This means that \(P(A \cap B) = P(A)P(B)\), \(P(B \cap C) = P(B)P(C)\), and \(P(A \cap C) = P(A)P(C)\). However, these events are not necessarily mutually independent.

Step 2: Analyze the Given Probability

It is given that \(P(A \cap B \cap C) = 0\). Since the probability of their intersection is zero, it indicates that the three events do not occur simultaneously. However, it does not specify the relationship between pairs of events.

Step 3: Use Complement and Conditional Probability

We need to find \(P[(\bar{A} \cap \bar{B}) \mid C]\). By the definition of conditional probability: \[ P[(\bar{A} \cap \bar{B}) \mid C] = \frac{P[(\bar{A} \cap \bar{B}) \cap C]}{P(C)} \] Since \(A \cap B \cap C = \emptyset\), returning \(P(A \cap B \mid C) = 0\). Thus, all probabilities involving additional constraints with \(A, B, C\) together become zero.

Step 4: Apply Inclusion-Exclusion and Independence

Using \(P[\bar{A} \cap \bar{B} \cap C] = P[\bar{A} \cap C] - P[(\bar{A} \cap B) \cap C]\) and knowing that \(A\) and \(B\) are independent of \(C\), it holds: \[ P(\bar{A} \cap C) = P(\bar{A})P(C), \quad P(\bar{B} \cap C) = P(\bar{B})P(C)\] As \(A\) and \(B\) are independent, \(P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B})\), upon simplifying with \(P(A) + P(\bar{A}) = 1\), \(P(B) + P(\bar{B}) = 1\).

Step 5: Derive the Final Solution

Because \(P((A \cap B) \cap C) = 0\) and events are pairwise independent, the intersection results in zero. Considering complement intersections remains, the solution becomes:\[ P[(\bar{A} \cap \bar{B}) \mid C] = P(\bar{A})P(\bar{B}) \]Thus, adding component probabilities for simplification shows: \(P(\bar{A}) + P(\bar{B})\). The correct choice is (d): \(P(\bar{A}) + P(\bar{B})\).

Key Concepts

Conditional Probability

Conditional probability is a crucial concept when dealing with dependent events. It's helpful to determine the likelihood of an event occurring, given the occurrence of another event. In mathematical terms, the conditional probability of an event A given another event C is expressed as:\[ P(A \mid C) = \frac{P(A \cap C)}{P(C)} \]The formula shows the ratio of the probability that both A and C occur, to the probability that C occurs. This formula is particularly useful when events are not independent of each other. For instance, in our problem, we want to find the conditional probability \(P[(\bar{A} \cap \bar{B}) \mid C]\). The key step is understanding how the probability of \((\bar{A} \cap \bar{B})\) changes given that C has occurred. Using this principle helps simplify complex probability problems by breaking them down into more manageable calculations.

Complement of Events

Understanding the complement of events is central to probability calculus. The complement of an event A, denoted by \(\bar{A}\), consists of all outcomes in the sample space that are not in A. Therefore, the probability of the complement of A is simply one minus the probability of A:\[ P(\bar{A}) = 1 - P(A) \]Working with complements helps simplify probability calculations, especially when we want to subtract probabilities or look for non-occurrences. For our exercise, because we know the events are not occurring simultaneously \((P(A \cap B \cap C) = 0)\), utilizing complements like \(\bar{A}\) and \(\bar{B}\) is effective. By doing this, we calculate how likely it is that neither A nor B happens given that C does occur.

Inclusion-Exclusion Principle

The inclusion-exclusion principle is a combinatorial technique used to calculate the probability of the union of events. It corrects the over-counting that occurs if simple addition of probabilities is used. For two events A and B, the principle states:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]This method extends straightforwardly to more than two events, adjusting for intersections at higher orders. In our given problem, since the events are pairwise independent, the principle helps in computing probabilities involving complements. By leveraging the independence property \((P(A \cap B) = P(A)P(B))\), we can simplify calculations for complementary events \((P(\bar{A} \cap \bar{B}))\). Lastly, pairing inclusion-exclusion with complement rules, we've efficiently determined that the solution simplifies to \(P(\bar{A}) + P(\bar{B})\).