Problem 170 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(\mathrm{A}\) and \(\mathrm{B}\) be two events such that \(p(\overline{A \cup B})=\frac{1}{6}, p(A \cap B)=\frac{1}{4}\) and \(p(\bar{A})=\frac{1}{4}\), where \(\bar{A}\) stands for the complement of the event \(A\). Then, the events \(A\) and \(B\) are (A) mutually exclusive and independent (B) equally likely but not independent (C) independent but not equally likely (D) independent and equally likely

Step-by-Step Solution

Answer

The events A and B are independent but not equally likely.

Step 1: Determine the Probability of A or B

Given that the probability of neither A nor B happening is \( p(\overline{A \cup B}) = \frac{1}{6} \). The probability of A or B happening is the complement of this, given by \( p(A \cup B) = 1 - p(\overline{A \cup B}) = 1 - \frac{1}{6} = \frac{5}{6} \).

Step 2: Use the Addition Rule for Probabilities

Using the formula \( p(A \cup B) = p(A) + p(B) - p(A \cap B) \), we substitute the known values: \( \frac{5}{6} = p(A) + p(B) - \frac{1}{4} \). Simplifying, \( p(A) + p(B) = \frac{5}{6} + \frac{1}{4} \).

Step 3: Solve for p(A) + p(B)

Convert fractions to a common denominator: \( \frac{5}{6} = \frac{10}{12} \) and \( \frac{1}{4} = \frac{3}{12} \). Thus, \( p(A) + p(B) = \frac{10}{12} + \frac{3}{12} = \frac{13}{12} \).

Step 4: Find the Probability of A

Since \( p(\bar{A}) = \frac{1}{4} \), it follows that \( p(A) = 1 - p(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4} \).

Step 5: Solve for the Probability of B

Using values found in steps 3 and 4, substitute \( p(A) = \frac{3}{4} = \frac{9}{12} \) into \( p(A) + p(B) = \frac{13}{12} \). Then \( \frac{9}{12} + p(B) = \frac{13}{12} \). Hence, \( p(B) = \frac{13}{12} - \frac{9}{12} = \frac{4}{12} = \frac{1}{3} \).

Step 6: Check for Independence

Events A and B are independent if \( p(A \cap B) = p(A) \times p(B) \). Calculate \( \frac{3}{4} \times \frac{1}{3} = \frac{1}{4} \). Since \( p(A \cap B) = \frac{1}{4} \) matches, A and B are independent.

Step 7: Check if A and B are Equally Likely

Events A and B are equally likely if \( p(A) = p(B) \). Since \( p(A) = \frac{3}{4} \) and \( p(B) = \frac{1}{3} \), they are not equally likely.

Key Concepts

Independence of Events

In probability theory, two events are said to be independent if the occurrence of one event does not affect the probability of the other event occurring. This is a critical concept, as it helps separate events that have no influence on each other. For events to be defined as independent, the following condition must be satisfied:

• If events A and B are independent, then the probability of both A and B happening together is equal to the product of their individual probabilities.

Mathematically, this can be expressed as: \[ p(A \cap B) = p(A) \times p(B) \]In the provided solution, events A and B were evaluated for independence. By calculating \( \frac{3}{4} \times \frac{1}{3} = \frac{1}{4} \), which equaled the given \( p(A \cap B) \), they confirmed independence. It shows that the likelihood of A happening in no way changes the probability of B occurring and vice versa, emphasizing the randomness and independence of these events.Understanding independence helps in determining the real influence or correlation one event may have over another. In practical terms, if you flip a coin and roll a die simultaneously, the result of the coin flip does not affect the outcome of the die roll. They are inherently independent events.

Mutually Exclusive Events

Mutually exclusive events are those that cannot happen at the same time. If one event occurs, it means that the other cannot. This is straightforward and helps in delineating events that have clear boundaries.Mathematically, mutually exclusive events, A and B, satisfy the following condition:

• \( p(A \cap B) = 0 \) - This means that there is zero probability of both events occurring together.

In the context of the exercise, if events A and B were mutually exclusive, the intersection of A and B would be zero, but since \( p(A \cap B) \) was \( \frac{1}{4} \), they are not mutually exclusive. Understanding mutually exclusive events assists in clarifying scenarios where one outcome inherently bars another. Consider a six-sided die; rolling a 3 and rolling a 5 are mutually exclusive. You're cannot roll a 3 and a 5 simultaneously with a single die roll. This concept helps organize event scenarios in probability situations, ensuring clarity when determining combined probabilities or comparing different outcomes.

Addition Rule for Probabilities

The addition rule for probabilities is a fundamental principle used when determining the probability of either one of two events happening. This rule is particularly helpful in understanding combined probabilities when events aren't mutually exclusive.The Addition Rule is formulated as:\[ p(A \cup B) = p(A) + p(B) - p(A \cap B) \]This rule includes the subtraction of the intersection of A and B to avoid double-counting in situations where events A and B can happen simultaneously.In the solution context, the formula was applied to solve for \( p(A) + p(B) \). Given that the union \( p(A \cup B) \) was \( \frac{5}{6} \) and \( p(A \cap B) = \frac{1}{4} \), the addition rule helped derive these values effectively, leading to a seamless calculation process.The addition rule is crucial for constructing a complete picture of event interactions, enabling the assessment of various probability scenarios with overlapping events. It ensures that probabilities are calculated accurately, laying a strong foundation for evaluating complex or real-world situations.