Problem 170 - A Complete Resource Book in Mathematics for JEE Main 2017

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 (C).

Step 1: Use the Complement Rule

We have \(p(\overline{A \cup B}) = \frac{1}{6}\). Using the complement rule, we find the probability of the union: \[ p(A \cup B) = 1 - p(\overline{A \cup B}) = 1 - \frac{1}{6} = \frac{5}{6}. \]

Step 2: Apply the Formula for Union of Two Events

The formula for the probability of the union of two events is \( p(A \cup B) = p(A) + p(B) - p(A \cap B) \). Substitute the known values: \[ \frac{5}{6} = p(A) + p(B) - \frac{1}{4}. \]

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

Rearrange the equation from Step 2 to solve for \(p(A) + p(B)\): \[ p(A) + p(B) = \frac{5}{6} + \frac{1}{4} = \frac{10}{12} + \frac{3}{12} = \frac{13}{12}. \]

Step 4: Use Given Probability of \(\bar{A}\) to Solve for \(p(A)\)

Since \(p(\bar{A}) = \frac{1}{4}\), we can find \(p(A)\) using the complement rule: \[ p(A) = 1 - p(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4}. \]

Step 5: Find \(p(B)\) from \(p(A) + p(B)\)

We already found that \(p(A) + p(B) = \frac{13}{12}\). Now, substitute \(p(A) = \frac{3}{4}\) to find \(p(B)\): \[ p(B) = \frac{13}{12} - \frac{3}{4} = \frac{13}{12} - \frac{9}{12} = \frac{4}{12} = \frac{1}{3}. \]

Step 6: Check Independence of Events

Events \(A\) and \(B\) are independent if \(p(A \cap B) = p(A) \cdot p(B)\). Calculate this: \[ p(A) \cdot p(B) = \frac{3}{4} \cdot \frac{1}{3} = \frac{3}{12} = \frac{1}{4}. \] \(p(A \cap B)\) is also given as \(\frac{1}{4}\). Since these are equal, \(A\) and \(B\) are independent.

Step 7: Check if Events are Equally Likely

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

Key Concepts

Conditional Probability

Conditional probability is a powerful concept that helps us understand the likelihood of an event, given that another event has already occurred. It's represented as \( p(A|B) \), which reads as "the probability of \( A \) given \( B \)." This is particularly useful in situations where we need to adjust our predictions based on new evidence or occurrences.

To calculate conditional probability, we use the formula:

• \( p(A|B) = \frac{p(A \cap B)}{p(B)} \)

If we know that event \( B \) has occurred, we can filter our outcome space to just those scenarios where \( B \) is true.

It's important to note that conditional probability is only defined when the event we are conditioning on \( B \) has a probability greater than zero \( p(B) > 0 \). Otherwise, the probability of \( B \) occurring is impossible, and thus, we can't make further assumptions about \( A \).

Independent Events

Two events are said to be independent if the occurrence or non-occurrence of one does not affect the probability of the other occurring. In other words, knowledge about one event happening provides no information about the other.

The test for independence is straightforward: two events \( A \) and \( B \) are independent if the equation \( p(A \cap B) = p(A) \cdot p(B) \) holds true.

This was verified in the step-by-step solution where \( p(A \cap B) \) was calculated to be equivalent to \( p(A) \cdot p(B) \), demonstrating that \( A \) and \( B \) are indeed independent events.

When events are independent:

• The conditional probability \( p(A|B) \) equals \( p(A) \)
• The same holds for \( B|A \) with \( p(B|A) = p(B) \)

These rules highlight how independence simplifies probability calculations, as you don’t need information from one event to predict the other.

Complement Rule

The complement rule is an exceptionally handy tool in probability theory. It provides an easy way to calculate the probability of an event not occurring. The complement of an event \( A \) is represented as \( \bar{A} \), and its probability is given by:

• \( p(\bar{A}) = 1 - p(A) \)

This takes advantage of the fact that the probabilities of all possible outcomes of an experiment must sum to one.

In practice, when given \( p(\bar{A}) \), the complement rule allows us to effortlessly flip it, finding \( p(A) \) by using the formula \( p(A) = 1 - p(\bar{A}) \).

In the exercise, this rule was applied to find \( p(A) \) when \( p(\bar{A}) \) was known. It's particularly useful when the probability of something not happening is easier to determine than the event itself, as seen in inverse-selected processes.