Problem 79 - A Complete Resource Book in Mathematics for JEE Main

Question

If two events \(A\) and \(B\) are such that \(P\left(A^{\prime}\right)=0.3, P(B)=\) \(0.4\) and \(P\left(A \cap B^{\prime}\right)=0.5\), then \(P\left(B / A \cup B^{\prime}\right)\) equals (A) \(\frac{3}{4}\) (B) \(\frac{5}{6}\) (C) \(\frac{1}{4}\) (D) \(\frac{3}{7}\)

Step-by-Step Solution

Answer

(C) \(\frac{1}{4}\)

Step 1: Calculate P(A)

Since we know \( P(A') = 0.3 \), we can find \( P(A) \) using the relation \( P(A) + P(A') = 1 \). Thus, \( P(A) = 1 - 0.3 = 0.7 \).

Step 2: Find P(A ∪ B')

We know that \( P(A \cup B') = P(A) + P(B') - P(A \cap B') \). Given that \( P(B') = 1 - P(B) = 1 - 0.4 = 0.6 \) and \( P(A \cap B') = 0.5 \), we can substitute these values to get \( P(A \cup B') = 0.7 + 0.6 - 0.5 = 0.8 \).

Step 3: Calculate P(B ∩ (A ∪ B'))

Since the problem asks for \( P(B / (A \cup B')) \), we first need \( P(B \cap (A \cup B')) \). We use the distributive law: \( P(B \cap (A \cup B')) = P((B \cap A) \cup (B \cap B')) \). We know \( P(B') = 0.6 \), so \( P(B \cap B') = 0 \) because the intersection is impossible. Hence, \( P(B \cap (A \cup B')) = P(B \cap A) \).

Step 4: Use total probability for P(B ⋂ A)

We know \( P(A) = 0.7 \) and \( P(A \cap B') = 0.5 \), so \( P(A \cap B) = P(A) - P(A \cap B') = 0.7 - 0.5 = 0.2 \). Hence, \( P(B \cap (A \cup B')) = 0.2 \).

Step 5: Compute P(B / (A ∪ B'))

The conditional probability \( P(B / (A \cup B')) \) is given by the formula \( \frac{P(B \cap (A \cup B'))}{P(A \cup B')} \). From Steps 2 and 4, we have \( P(B \cap (A \cup B')) = 0.2 \) and \( P(A \cup B') = 0.8 \). Thus, \( P(B / (A \cup B')) = \frac{0.2}{0.8} = \frac{1}{4} \).

Key Concepts

Conditional Probability

When dealing with events in probability, we often want to know the likelihood of an event occurring given that another event has already occurred. This is known as conditional probability. It helps us refine the probability assessment based on additional information. Consider two events, A and B. The conditional probability of B given A, represented as \( P(B|A) \), is calculated by:\[P(B|A) = \frac{P(A \cap B)}{P(A)}\]This formula means you're finding the probability of both A and B happening and dividing it by the probability of A happening.

• If \( A \) has occurred, you use it to update the likelihood of \( B \).
• Conditional probability gives insight into dependent events.
• It helps in various fields such as machine learning, risk assessment, and decision-making.

Union of Events

The union of two events, denoted by \( A \cup B \), consists of outcomes that are in either event A or B, or in both. It's useful for combining probabilities from multiple scenarios. The formula for the probability of the union of two events is:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]This equation ensures that any overlap between A and B is subtracted once, as it's counted twice when adding the probabilities of A and B separately.

• Union is essential for events that can happen simultaneously.
• You need to consider the intersection to avoid double-counting.
• Union is used in fields like insurance to calculate total risks.

Intersection of Events

The intersection of two events, written as \( A \cap B \), contains outcomes that are common to both events A and B. It is the "and" operation in set theory, where both conditions must be true. The probability of the intersection is pivotal in understanding the overlap of events.When determining probabilities, the formula is:\[P(A \cap B) = P(B \cap A)\]This formula underscores that intersection is commutative. The probability of A and B occurring is the same as B and A. Knowing how to calculate intersections is crucial for both dependent and independent events.

• Intersection helps in scenarios where multiple criteria must be satisfied.
• It is a core concept in probability theory and statistical analysis.
• Intersection aids in understanding real-world problems like traffic flow analysis and system reliability.