Problem 7 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A\) and \(B\) be two independent events such that \(P(A)=\frac{1}{3}\) and \(P(B)=\frac{1}{6} .\) Then, which of the following is \(\begin{aligned}&\text { TRUE? } & \text { [Jan. 8, 2020 (I)] }\end{aligned}\) (a) \(P(A / B)=\frac{2}{3}\) (b) \(P\left(A / B^{\prime}\right)=\frac{1}{3}\) (c) \(P\left(A^{\prime} / B^{\prime}\right)=\frac{1}{3}\) (d) \(P(A /(A \cup B))=\frac{1}{4}\)

Step-by-Step Solution

Answer

Option (b) is true.

Step 1: Understand Independence

In probability, two events \( A \) and \( B \) are independent if the occurrence of one does not affect the probability of the other. Therefore, \( P(A \mid B) = P(A) \). This means the conditional probability of \( A \) given \( B \) is the same as \( P(A) \).

Step 2: Analyze Option (a)

Option (a) states \( P(A \mid B) = \frac{2}{3} \). However, since \( A \) and \( B \) are independent, \( P(A \mid B) = P(A) = \frac{1}{3} \), not \( \frac{2}{3} \). So, (a) is false.

Step 3: Analyze Option (b)

Option (b) states \( P(A \mid B') = \frac{1}{3} \). For independent events, the probability of \( A \) given the complement of \( B \), \( B' \), is still the probability of \( A \). Thus, \( P(A \mid B') = P(A) = \frac{1}{3} \). So, (b) is true.

Step 4: Analyze Option (c)

Option (c) is \( P(A' \mid B') = \frac{1}{3} \). However, since \( P(A') = 1 - P(A) = \frac{2}{3} \), and because \( A \) and \( B \) are independent, \( P(A' \mid B') = P(A') = \frac{2}{3} \), not \( \frac{1}{3} \). Hence, (c) is false.

Step 5: Analyze Option (d)

Option (d) considers \( P(A \mid A \cup B) = \frac{1}{4} \). Using the formula for conditional probability, \( P(A \mid A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)} \). Calculate \( P(A \cap B) = P(A)P(B) = \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{18} \).\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{6} - \frac{1}{18} = \frac{6}{18} + \frac{3}{18} - \frac{1}{18} = \frac{8}{18} = \frac{4}{9} \).Hence, \( P(A \mid A \cup B) = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{1}{3} \times \frac{9}{4} = \frac{3}{4} \), not \( \frac{1}{4} \). So, (d) is false.

Key Concepts

Conditional Probability

Conditional probability is a crucial concept that helps you understand the likelihood of an event given that another event has already occurred. For events \(A\) and \(B\), the conditional probability of \(A\) given \(B\), denoted by \(P(A \mid B)\), is found using the formula:

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]
However, when \(A\) and \(B\) are independent events, meaning the occurrence of \(A\) does not affect \(B\) and vice-versa, the formula simplifies. This means \(P(A \mid B) = P(A)\), since the occurrence of \(B\) does not impact the likelihood of \(A\).

In the provided exercise, since \(A\) and \(B\) are independent, the correct conditional probabilities reflect their actual probabilities as separate occurrences. Hence, options like \(P(A \mid B)\) and \(P(A \mid B')\) leverage the independence property to simply become \(P(A) = \frac{1}{3}\). Understanding this makes it easier to identify conditional probabilities for independent events.

Complementary Events

Complementary events occur when you have an event and its opposite. If \(A\) is an event, then its complement \(A'\) represents all the outcomes that are not in \(A\). For any event \(A\), the sum of the probabilities of \(A\) and its complement is always 1, expressed by the equation:

\[ P(A') = 1 - P(A) \]
This basic principle helps solve problems by focusing on what remains after the event has occurred. Thus, in situations involving the complements of events, like \(P(A' \mid B')\), the key is understanding that the sum of an event and its complement within a sample space always equals the total probability — anything outside the likelihood of \(A\) is part of \(A'\).

In the example, \(P(A' \mid B')\) misapplies the concept, relying instead on \(P(A')\) alone. For independent events, \(P(A' \mid B') = P(A') \). This is because the occurrence of \(B\) or its complement \(B'\) does not influence the probability of \(A'\), sticking to the independent principle.

Probability of Union of Events

The probability of the union of events \(A\) and \(B\), denoted \(P(A \cup B)\), represents the likelihood that either \(A\) or \(B\) or both occur. The key formula for calculating the union probability is:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Subtracting \(P(A \cap B)\) ensures that any overlap where both events occur simultaneously is not counted twice. When the events are independent, \(P(A \cap B)\) is simply \(P(A) \cdot P(B)\), since the independence implies no extra influence on occurrence probability by one event on the other.

In the exercise, determining \(P(A \cup B)\) involves both calculation of the individual probabilities and the intersection, which adjusts for mutual occurrences. Grasping the concept of union probability is vital for correctly combining individual event probabilities within a given sample space.