Problem 158 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(A\) and \(B\) 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 complement of event \(A\). Then events \(A\) and \(B\) are (A) equally likely and mutually exclusive (B) equally likely but not independent (C) independent but not equally likely (D) mutually exclusive and independent

Step-by-Step Solution

Answer

Events A and B are independent but not equally likely.

Step 1: Understanding Complement of Events

First, note that the probability of the complement of an event, for instance, \(A \cup B\), is given by\[P(\overline{A \cup B}) = 1 - P(A \cup B)\]Since \(P(\overline{A \cup B}) = \frac{1}{6}\), therefore\[P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}\]

Step 2: Calculate Probability of A

Since \(P(\bar{A}) = \frac{1}{4}\), we find \(P(A)\) as follows:\[P(A) = 1 - P(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4}\]

Step 3: Apply the formula for Union of Two Events

We know the formula for probability of union of two events:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]Substituting known values:\[\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}\]\[\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}\]

Step 4: Simplify for P(B)

Let's simplify the expression to solve for \(P(B)\):\[\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}\]\[\frac{5}{6} = \frac{1}{2} + P(B)\]\[P(B) = \frac{5}{6} - \frac{1}{2}\]\[P(B) = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}\]

Step 5: Check for Independence

For independence, we need to check if\[P(A \cap B) = P(A) \cdot P(B)\]Substituting the values:\[\frac{1}{4} = \frac{3}{4} \cdot \frac{1}{3}\]\[\frac{1}{4} = \frac{3}{12} = \frac{1}{4}\]Yes, they are equal, thus events \(A\) and \(B\) are independent.

Step 6: Analyze Conditions from Options

- Mutually exclusive events satisfy \(P(A \cap B) = 0\). Here \(P(A \cap B) = \frac{1}{4}\), so they are not mutually exclusive.- Equally likely would mean \(P(A) = P(B)\); \(\frac{3}{4} eq \frac{1}{3}\).- Therefore, the correct option is (C) independent but not equally likely.

Key Concepts

Independent Events

To determine whether two events are independent, we check for a specific condition. Two events, say event \(A\) and event \(B\), are considered independent if the occurrence of one does not affect the occurrence of the other. This can be mathematically represented as \[ P(A \cap B) = P(A) \times P(B) \]
If this equation holds true, we conclude that the events are independent.Let's take an example from the exercise. Imagine events \(A\) and \(B\) having the following probabilities given:

• \(P(A) = \frac{3}{4}\)
• \(P(B) = \frac{1}{3}\)
• \(P(A \cap B) = \frac{1}{4}\)

Using the formula above, you can substitute the values:\[ \frac{3}{4} \times \frac{1}{3} = \frac{1}{4} \]This confirms \(A\) and \(B\) are indeed independent. Remember, independence implies that one event happening does not change the probability of the other occurring.

Mutually Exclusive Events

Understanding mutually exclusive events is key to many probability problems. Mutually exclusive events are defined as events that cannot occur at the same time. In simpler words, if one event occurs, the other cannot. This means the intersection, or the overlap, of the events is zero: \[ P(A \cap B) = 0 \]Let's revisit our exercise example. If \(P(A \cap B) = \frac{1}{4}\), then \(A\) and \(B\) are not mutually exclusive.
Had they been mutually exclusive, \(P(A \cap B)\) would have equaled zero.Here’s a practical example: if you have a single coin toss, getting heads and tails simultaneously is impossible. That’s mutual exclusivity—one excludes the other. Always remember, for mutually exclusive events, knowing one occurs means the other will not occur, which is a critical difference from independence.

Complementary Events

Complementary events are two outcomes that cover all possibilities of a particular probability scenario. They are like two halves of a whole, meaning if one happens, the other doesn’t, and the probability of all outcomes adds up to 1.For any event \(A\), the complement is denoted by \(\bar{A}\), representing all outcomes not in \(A\). Thus, their probabilities satisfy:\[ P(A) + P(\bar{A}) = 1 \]
For instance, in the exercise, \(P(\bar{A}) = \frac{1}{4}\), giving:\[ P(A) = 1 - \frac{1}{4} = \frac{3}{4} \]The sum \(P(A) + P(\bar{A})\) verifies the completeness of all possible outcomes.Consider a die roll: the event \(A\) could be landing a six. Thus, \(\bar{A}\) includes everything from one to five. Since these cover all possibilities of a fair die, they’re complementary.Always use the complement for easy calculations—if you know the chance of something not happening, you instantly know the chance of it happening simply by subtracting from one.