Problem 5 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(E^{C}\) denote the complement of an event \(E\). Let \(E_{1}, E_{2}\) and \(\mathrm{E}_{3}\) be any pairwise independent events with \(\mathrm{P}\left(\mathrm{E}_{1}\right)>0\) and \(\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3}\right)=0\). Then \(\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}} \cap \mathrm{E}_{3}^{\mathrm{C}} / \mathrm{E}_{1}\right)\) is equal to: \([\) Sep. \(\mathbf{0 2}, \mathbf{2 0 2 0}(\mathrm{II})]\) (a) \(\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right)\) (b) \(\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)\) (c) \(\mathrm{P}\left(\mathrm{E}_{3}\right)-\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)\) (d) \(\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}\right)\)

Step-by-Step Solution

Answer

The answer is option (d): \( \mathrm{P}(E_3^C) - \mathrm{P}(E_2) \).

Step 1: Understand the Problem

We are given three pairwise independent events \( E_1, E_2, \) and \( E_3 \) with certain conditions. We need to find the probability of the events \( E_2^C \) and \( E_3^C \) occurring given \( E_1 \), represented as \( \mathrm{P}(E_2^C \cap E_3^C / E_1) \). The conditions state that \( \mathrm{P}(E_1 \cap E_2 \cap E_3) = 0 \).

Step 2: Use Conditional Probability Definition

By definition, the conditional probability \( \mathrm{P}(A / B) \) is given by \( \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \) given that \( \mathrm{P}(B) > 0 \). Apply this definition to find \( \mathrm{P}(E_2^C \cap E_3^C / E_1) \), which is \( \frac{\mathrm{P}(E_2^C \cap E_3^C \cap E_1)}{\mathrm{P}(E_1)} \).

Step 3: Analyze the Complement of Events Intersection

We need to find \( \mathrm{P}(E_2^C \cap E_3^C \cap E_1) \). Notice that \( E_2^C \cap E_3^C \cap E_1 \) means \( E_1 \) occurs but neither \( E_2 \) nor \( E_3 \) occur. Thus, the events \( E_2^C \) and \( E_3^C \) must happen simultaneously with \( E_1 \).

Step 4: Use Pairwise Independence

Since the events are pairwise independent, we have \( \mathrm{P}(E_1 \cap E_2) = \mathrm{P}(E_1) \cdot \mathrm{P}(E_2) \) and similar relations for other pairs. For \( E_2^C \) and \( E_3^C \), we need to compute their simultaneous probability with \( E_1 \).

Step 5: Apply Given Condition

We are given that \( \mathrm{P}(E_1 \cap E_2 \cap E_3) = 0 \). This implies that whenever both \( E_2 \text{and} \ E_3 \) occur, \( E_1 \) cannot be true. Hence, the only contributing scenario to \( E_1 \) is when both \( E_2 \) and \( E_3 \) do not occur (i.e., their complements occur). Thus, \( \mathrm{P}(E_2^C \cap E_3^C \cap E_1) = \mathrm{P}(E_1 \cap E_2^C \cap E_3^C) \).

Step 6: Compute Probability Expression

Using pairwise independence and substituting the complement event, we find \( \mathrm{P}(E_1 \cap E_2^C) = \mathrm{P}(E_1) \cdot (1 - \mathrm{P}(E_2)) \) and similarly \( \mathrm{P}(E_1 \cap E_3^C) = \mathrm{P}(E_1) \cdot (1 - \mathrm{P}(E_3)) \). Use these results in the expression derived in Step 2, considering independence between the terms.

Step 7: Simplification and Final Result

Combine the probability rules to get \( \mathrm{P}(E_2^C \cap E_3^C / E_1) = 1 - \mathrm{P}(E_2) - \mathrm{P}(E_3) \) and notice from the options, the closest match given the combinations is \( \mathrm{P}(E_3^C) - \mathrm{P}(E_2) \). Thus the answer corresponds to option (d).

Key Concepts

Conditional Probability

Understanding conditional probability is essential as it allows us to calculate the probability of an event occurring given that another event has occurred. Formally, if we have two events, say \( A \) and \( B \), the conditional probability of \( A \) given \( B \) is denoted as \( \mathrm{P}(A / B) \) and is calculated using the formula: \[ \mathrm{P}(A / B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \] This formula requires that \( \mathrm{P}(B) > 0 \) because you cannot condition on an event that never happens. In practical terms, if you know \( B \) has occurred, then the universe of possible outcomes is reduced to just those outcomes where \( B \) is true. The probability of \( A \) given \( B \), then is the probability that both \( A \) and \( B \) occur (the intersection), compared to the probability that \( B \) itself occurs.

Complement of an Event

The complement of an event is all that is not included in the event. For any event \( E \), its complement is denoted by \( E^C \). The idea is that if \( E \) does not happen, then its complement \( E^C \) must occur. It is expressed in probability terms as:\[ \mathrm{P}(E^C) = 1 - \mathrm{P}(E) \] This equation arises because the total probability of all potential outcomes (\( E \) and \( E^C \)) in a probability space always adds up to 1.Complements are helpful for calculating probabilities indirectly. Sometimes it is easier to compute \( \mathrm{P}(E^C) \) first, then use it to find \( \mathrm{P}(E) \). This concept is extremely useful in solving complicated problems involving multiple events, as it allows you to focus on what is not happening and derive information from there.

Probability Rules

Probability rules are the foundational principles that guide how probabilities are computed and manipulated. Knowing these rules is crucial to solve probability problems effectively:

• Addition Rule: This rule states that for any two mutually exclusive events \( A \) and \( B \), the probability that either \( A \) or \( B \) occurs is the sum of their probabilities: \( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) \). For non-mutually exclusive events, you subtract the intersection: \( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) \).
• Multiplication Rule: For independent events, the probability that both \( A \) and \( B \) occur is the product of their individual probabilities: \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \cdot \mathrm{P}(B) \).

Moreover, these rules are used along with complements and conditional probability to analyze and solve complex probability scenarios, as they provide a clear structure for breaking down problems.

Independence in Probability

Independence is a key concept in probability that simplifies the calculation of probabilities of simultaneous events. Two events \( A \) and \( B \) are considered independent if the occurrence of one does not affect the probability of the other occurring. In formula terms, this is expressed as:\[ \mathrm{P}(A \cap B) = \mathrm{P}(A) \cdot \mathrm{P}(B) \] If events are independent, then knowing that \( A \) has occurred does not change the probability of \( B \) occurring.In the case of pairwise independent events \( E_1, E_2, \) and \( E_3 \), as the problem describes, this property holds true for each pair, without implying that all three together are independent. Pairwise independence simplifies calculations because you can treat each pair separately without considering additional dependence among other events. This distinction can be crucial, especially when evaluating the collective behaviour of such events, such as in intersections or unions.