Problem 17 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(A\) and \(B\) be two events such that the probability that exactly one of them occurs is \(\frac{2}{5}\) and the probability that \(A\) or \(B\) occurs is \(\frac{1}{2}\), then the probability of both of them occur together is: \(\quad\) [Jan. 8, 2020 (II)] (a) \(0.02\) (b) \(0.20\) (c) \(0.01\) (d) \(0.10\)

Step-by-Step Solution

Answer

The correct answer is (d) 0.10.

Step 1: Define the given probabilities

We are told that the probability that exactly one of the events occurs is \( \frac{2}{5} \). This can be represented as:\[ P(A \text{ and not } B) + P(B \text{ and not } A) = \frac{2}{5} \]Additionally, we are given that the probability that either event occurs is \( \frac{1}{2} \), which is:\[ P(A \cup B) = \frac{1}{2} \]

Step 2: Relationship between probabilities

Recall that the probability of the union of events can be expressed using the formula:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]We also know that:\[ P(A \cup B) = P(A \text{ and not } B) + P(B \text{ and not } A) + P(A \cap B) \]

Step 3: Substitute and solve for intersection probability

Substituting the known values into the second expression from the relationship in Step 2, we have:\[ \frac{1}{2} = \frac{2}{5} + P(A \cap B) \]Subtract \( \frac{2}{5} \) from both sides to solve for \( P(A \cap B) \):\[ P(A \cap B) = \frac{1}{2} - \frac{2}{5} \]

Step 4: Calculate the result

Convert the fractions to have a common denominator:\[ \frac{1}{2} = \frac{5}{10}, \quad \frac{2}{5} = \frac{4}{10} \]So,\[ P(A \cap B) = \frac{5}{10} - \frac{4}{10} = \frac{1}{10} \]

Step 5: Finalize and choose answer

The probability that both events \(A\) and \(B\) occur together is \(\frac{1}{10} = 0.10\). Therefore, the correct answer is option (d).

Key Concepts

Events in Probability

Events in probability are outcomes or combinations of outcomes from a random experiment. An event might be simple or compound. For instance, in rolling a six-sided die, obtaining a number more than four is a simple event.
Whether flipping a coin, rolling a die, or drawing a card, each of these involves events. Their probabilities help in predicting the likelihood/occurrence of future events. These probabilities are central to the study of probability theory.

• Simple Event: An individual outcome.
• Compound Event: A combination of two or more simple events.

In the original exercise, two events denoted as \(A\) and \(B\) were analyzed. We calculated how certain events like exactly one event happening, or both events occurring could be determined through given conditions.

Probability Formulas

Probability formulas allow us to quantitatively express the likelihood that a particular event will occur. A common formula is the **probability of the union of two events**:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
This formula calculates the probability that either event \(A\), event \(B\), or both happen. In our exercise, this was critical.
The given formula breaks down as follows:

• \(P(A)\) is the probability of event \(A\) alone.
• \(P(B)\) is the probability of event \(B\) alone.
• \(P(A \cap B)\) is the probability of both events occurring together.

By substituting known probabilities in these formulas, we calculated the unknowns, such as finding \(P(A \cap B)\), the probability of both events. This exercise was an excellent application of these key probability principles.

Union and Intersection of Events

When discussing probability, two critical concepts are the union and intersection of events. Understanding them is key to solving many probability problems.

• Union of Events \(A \cup B\) includes all outcomes in either \(A\) or \(B\) or both. It is essentially the combination of the two sets.
• Intersection of Events \(A \cap B\) involves outcomes present in both events. It signifies the overlap/common outcomes of \(A\) and \(B\).

In the context of our problem, knowing that the probability of these overlaps or combinations was crucial. We first determined the probability of exactly one event occurring, and then turned to find \(P(A \cap B)\), which is the intersection.
Solving the exercise involved subtracting the known probability of exactly one event from the probability of the union of events. This system ensured accuracy, demonstrating how these concepts intertwine to reveal the underlying structure of probability calculations.