Problem 27 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{E}\) and \(\mathrm{F}\) be two independent events. The probability that both \(\mathrm{E}\) and \(\mathrm{F}\) happen is \(\frac{1}{12}\) and the probability that neither E nor F happens is \(\frac{1}{2}\), then a value of \(\frac{\mathrm{P}(\mathrm{E})}{\mathrm{P}(\mathrm{F})}\) is : [Online April 9, 2017] (a) \(\frac{4}{3}\) (b) \(\frac{3}{2}\) (c) \(\frac{1}{3}\) (d) \(\frac{5}{12}\)

Step-by-Step Solution

Answer

The value of \( \frac{\mathrm{P}(E)}{\mathrm{P}(F)} \) is \( \frac{4}{3} \).

Step 1: Understanding the Problem

We are given that two events E and F are independent. The probability of both events happening is given as \( \mathrm{P}(E \cap F) = \frac{1}{12} \). Furthermore, the probability of neither event happening is \( \mathrm{P}(E' \cap F') = \frac{1}{2} \). We need to determine \( \frac{\mathrm{P}(E)}{\mathrm{P}(F)} \).

Step 2: Using the Independence Property

Since E and F are independent events, the probability that both happen is the product of their individual probabilities: \( \mathrm{P}(E \cap F) = \mathrm{P}(E) \cdot \mathrm{P}(F) \). From the problem statement, \( \mathrm{P}(E) \cdot \mathrm{P}(F) = \frac{1}{12} \).

Step 3: Express Probability of Neither Happening

The probability that neither E nor F happens is given by: \( \mathrm{P}(E' \cap F') = \mathrm{P}(E') \cdot \mathrm{P}(F') = (1 - \mathrm{P}(E))(1 - \mathrm{P}(F)) = \frac{1}{2} \).

Step 4: Set Up Equations

We now have two equations: 1) \( \mathrm{P}(E) \cdot \mathrm{P}(F) = \frac{1}{12} \) and 2) \((1 - \mathrm{P}(E))(1 - \mathrm{P}(F)) = \frac{1}{2} \).

Step 5: Solve Simultaneously

From Equation 1, express \( \mathrm{P}(F) = \frac{1}{12 \cdot \mathrm{P}(E)} \). Substitute this into Equation 2 to solve for \( \mathrm{P}(E) \).

Step 6: Solve for P(E) and P(F)

Substituting \( \mathrm{P}(F) = \frac{1}{12 \cdot \mathrm{P}(E)} \) in \((1 - \mathrm{P}(E))(1 - \frac{1}{12 \cdot \mathrm{P}(E)}) = \frac{1}{2} \) gives a quadratic equation. Solve this quadratic equation to find \( \mathrm{P}(E) \) and then subsequently \( \mathrm{P}(F) \).

Step 7: Calculate the Ratio

After finding \( \mathrm{P}(E) \) and \( \mathrm{P}(F) \), calculate the ratio \( \frac{\mathrm{P}(E)}{\mathrm{P}(F)} \) to arrive at \( \frac{4}{3} \).

Key Concepts

Probability Calculations

Probability calculations are fundamental in understanding how likely events are to occur. In this exercise, we are given information about two independent events, E and F.
To calculate the probability that both events occur, we use their individual probabilities. For independent events, this is done by multiplying their probabilities together:

• \( \mathrm{P}(E \cap F) = \mathrm{P}(E) \cdot \mathrm{P}(F) \)

In our problem, \( \mathrm{P}(E \cap F) = \frac{1}{12} \), meaning the product of the probabilities of E and F occurring is \( \frac{1}{12} \). This straightforward calculation is a crucial step in probability problems. It helps in breaking down complicated probability questions into simpler, manageable parts.
Furthermore, the calculation of probability when neither event occurs uses the complementary probabilities:

• \( \mathrm{P}(E') = 1 - \mathrm{P}(E) \)
• \( \mathrm{P}(F') = 1 - \mathrm{P}(F) \)
• \( \mathrm{P}(E' \cap F') = (1 - \mathrm{P}(E))(1 - \mathrm{P}(F)) \)

This approach allows us to set up equations that are essential for solving the exercise. It emphasizes the importance of understanding and applying basic probability principles.

Independent Events in Mathematics

In mathematical terms, independent events are those where the occurrence of one event does not affect the occurrence of the other. This is a pivotal concept in probability theory.
Independent events demonstrate a key relationship:

• The probability that both independent events E and F occur simultaneously is simply the product of their individual probabilities.
• This is expressed as \( \mathrm{P}(E \cap F) = \mathrm{P}(E) \cdot \mathrm{P}(F) \).

Such properties simplify many probability calculations by allowing the determination of joint probabilities through multiplication.
Understanding independence is especially useful in problems where events have no influence over each other, ensuring that the probabilities can be dealt independently. This characteristic becomes particularly handy when dealing with complex probability scenarios or when analyzing statistical experiments involving multiple random events.
In the given exercise, the relationship persists and guides us through setting expressions that eventually solve the problem. Understanding this independent nature is key to unlocking the solution.

Mathematical Problem Solving

Solving mathematical problems often involves translating written problems into equations. This exercise is a classic example of how we can use known mathematical principles to solve for unknowns.
To tackle the problem given, we set up two equations from the problem statement:

• \( \mathrm{P}(E) \cdot \mathrm{P}(F) = \frac{1}{12} \)
• \((1 - \mathrm{P}(E))(1 - \mathrm{P}(F)) = \frac{1}{2} \)

From here, solving requires substitution and simplification. By expressing \( \mathrm{P}(F) \) from the product equation, it allows substitution into the second equation. Solving this algebraically results in a quadratic equation.
Such quadratic solutions often yield two potential answers, requiring interpretation to select the plausible one. In our example, these steps eventually lead us to find that \( \mathrm{P}(E) \) and \( \mathrm{P}(F) \) can be resolved, allowing us to compute the desired ratio \( \frac{\mathrm{P}(E)}{\mathrm{P}(F)} \).
The process highlights not just technical proficiency but the art of mathematical deduction. It's about breaking down problems into simpler known concepts and reformulating them into solvable formats. This skill is invaluable in navigating complex mathematical landscapes.