Problem 12 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If two events \(A\) and \(B\) are such that \(P\left(A^{c}\right)=0.3\), \(P(B)=0.4\) and \(P\left(A B^{c}\right)=0.5\), then \(P\left[B /\left(A \cup B^{c}\right)\right]\) is equal to (a) \(1 / 2\) (b) \(1 / 3\) (c) \(1 / 4\) (d) None of these

Step-by-Step Solution

Answer

The answer is option (d) None of these.

Step 1: Find P(A)

We know that the probability of an event not occurring is given by the complement of that event. This means if \( P(A^c) = 0.3 \), then \( P(A) = 1 - P(A^c) = 1 - 0.3 = 0.7 \).

Step 2: Identify Known Probabilities

We are given \( P(B) = 0.4 \) and \( P(A B^c) = 0.5 \). Now we need the probability \( P[A \cup B^c] \) for the next steps.

Step 3: Use Complementary Rule to Find P(B^c)

Since the probability of an event and its complement add up to 1, we find \( P(B^c) = 1 - P(B) = 1 - 0.4 = 0.6 \).

Step 4: Calculate P(A \cup B^c)

Since \( P(A \cup B^c) = P(A) + P(B^c) - P(A B^c) \), substitute the known values to get \( P(A \cup B^c) = 0.7 + 0.6 - 0.5 = 0.8 \).

Step 5: Use Conditional Probability Formula

We need to find \( P[B / (A \cup B^c)] \). The formula is: \[ P[B / (A \cup B^c)] = \frac{P[B \cap (A \cup B^c)]}{P(A \cup B^c)} \].

Step 6: Calculate P[B \cap (A \cup B^c)]

Since \( (A \cup B^c)^c = A^c \cap B \), we calculate \( P[(A \cup B^c)^c] = P(A^c \cap B) = P(B) - P(A B^c) = 0.4 - 0.5 = 0.1 \). Hence, \( P[B \cap (A \cup B^c)] = P(B) - P(A^c \cap B) = 0.4 - 0.1 = 0.3 \).

Step 7: Substitute Back into Formula

Substitute the values into the formula: \[ P[B / (A \cup B^c)] = \frac{0.3}{0.8} = \frac{3}{8} \].

Step 8: Find Closest Answer Option

Comparing \( \frac{3}{8} \) with the given options, none exactly match. Thus, the answer is option (d) None of these.

Key Concepts

Event Complement

In probability theory, every event has a counterpart known as its "complement." The complement of an event consists of all the possible outcomes that are not part of the event itself. If we denote an event by \( A \), its complement is usually denoted as \( A^c \). The fundamental principle here is that the sum of the probability of an event and the probability of its complement must equal 1.

For instance, consider an event \( A \) with \( P(A^c) = 0.3 \). This means that the probability of \( A \) not occurring is 0.3. According to the complement rule, the probability of the event \( A \) occurring would be \( P(A) = 1 - P(A^c) = 0.7 \).

Understanding complements can be incredibly helpful, especially when it’s easier to compute probabilities of things not happening rather than what is happening. This technique can simplify calculations and give a deeper insight into probability scenarios.

Probability of Union

The probability of a union of two events, denoted as \( P(A \cup B) \), refers to the likelihood of either event \( A \), event \( B \), or both occurring. 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 formula accounts for the fact that simply adding the probabilities of \( A \) and \( B \) would result in double-counting the intersection where both events occur.

In the given exercise, to find \( P(A \cup B^c) \), we utilize the complement \( B^c \), which denotes the event that \( B \) does not happen. Thus, the probability formula becomes \( P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) \). By substituting known values, these calculations simplify the task of finding the probability of either \( A \) or \( B^c \) occurring.

Probability of Intersection

The probability of the intersection of two events, expressed as \( P(A \cap B) \), signifies the chance that both events \( A \) and \( B \) occur simultaneously. This is a crucial concept in understanding how events overlap and influence each other in probability.

To calculate the probability of the intersection of two events, often given other probabilities, we can rearrange or utilize known complementary events. For example, consider leveraging \( P(A \cap B^c) \) in place of direct union calculations.

In strategies like the exercise, the relation between the union and complement or intersection helps solve for unknowns: \( P[(A \cup B^c)^c] = P(A^c \cap B) \). This equation can assist in uncovering values like \( P(A^c \cap B) \) by performing operations on known probabilities, helping to solve the structure of conditional probability scenarios effectively. Hence, mastering it allows for a deeper comprehension of how compounds of probability function together.