Problem 9 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is \(\frac{1}{4}\) If the probability that at most two machines will be out of service on the same day is \(\left(\frac{3}{4}\right)^{3} k\), then \(k\) is equal to: [Jan. 7, 2020 (II)] (a) \(\frac{17}{8}\) (b) \(\frac{17}{4}\) (c) \(\frac{17}{2}\) (d) 4

Step-by-Step Solution

Answer

The value of \( k \) is \( \frac{17}{4} \), which corresponds to option (b).

Step 1: Understanding the Problem

We are given a probability problem involving machines being out of service. The individual probability of a machine being out of service is \( \frac{1}{4} \), and the problem asks for the probability that at most two of these machines are out of service, which is given as \( \left( \frac{3}{4} \right)^3 k \). We need to find the value of \( k \).

Step 2: Defining the Probability Framework

Let's define our probability model: This situation follows a binomial distribution where the number of trials is 5 (since there are 5 machines), the probability of success (a machine being out of service) is \( p = \frac{1}{4} \), and the probability of failure (a machine not being out of service) is \( q = 1 - p = \frac{3}{4} \).

Step 3: Calculating the Probability Distribution

For at most two machines to be out of service, we need the probabilities of 0, 1, and 2 machines being out of service and sum them up. These events can be expressed using the binomial probability formula: \( P(X = k) = \binom{n}{k} p^k q^{n-k}\), where \( n = 5 \), \( p = \frac{1}{4} \), \( q = \frac{3}{4} \).

Step 4: Probability of 0 Machines Out of Service

Calculate the probability of 0 machines being out of service: \( \binom{5}{0} \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^5 = 1 \times 1 \times \left( \frac{3}{4} \right)^5 = \left( \frac{243}{1024} \right) \).

Step 5: Probability of 1 Machine Out of Service

Calculate the probability of 1 machine being out of service: \( \binom{5}{1} \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^4 = 5 \times \frac{1}{4} \times \frac{81}{256} = \frac{405}{1024} \).

Step 6: Probability of 2 Machines Out of Service

Calculate the probability of 2 machines being out of service: \( \binom{5}{2} \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^3 = 10 \times \frac{1}{16} \times \frac{27}{64} = \frac{270}{1024} \).

Step 7: Summing Probabilities of At Most 2 Machines Out

Sum the probabilities of 0, 1, and 2 machines being out of service: \( P(X \leq 2) = \frac{243}{1024} + \frac{405}{1024} + \frac{270}{1024} = \frac{918}{1024} = \left( \frac{3}{4} \right)^3 \cdot k \).

Step 8: Equating and Solving for k

We know \( \left( \frac{3}{4} \right)^3 = \frac{27}{64} \). Therefore, we have \( \frac{918}{1024} = \frac{27}{64} \cdot k \). Multiplying both sides by \( 64 \) gives \( 918 \times \frac{64}{1024} = 27k \). Simplifying, \( k = \frac{918 \times 64}{1024 \times 27} = \frac{17}{4} \).

Step 9: Identify the Correct Option

Check the calculated value of \( k = \frac{17}{4} \) against the given options. This matches option (b).

Key Concepts

Probability Theory

Probability theory is the branch of mathematics concerned with the analysis of random phenomena. At its core, probability theory involves determining how likely events are to occur. In the context of our exercise, we can explore this through the probability of machines being out of service at a workshop.

In probability:

• Each machine independently has a 1 in 4 chance of being out of service each day.
• The probability of an event not happening is 1 minus the probability of it happening. Here, that makes the chance of a machine functioning correctly \( \frac{3}{4} \).
• The sum of probabilities for all possible outcomes in a given scenario equals 1.

Understanding these basic principles helps you calculate complex probabilities, as seen in this exercise, where various outcomes for machines being either in or out of service are evaluated.

Mathematical Problem Solving

Solving mathematical problems often involves breaking a problem down into smaller, manageable steps. This exercise highlights the importance of:

• Establishing a clear problem statement - here it is to find the probability of at most two machines being out of service and calculating the constant \( k \).
• Identifying known elements - in this instance, the probability associated with each machine's operational status.
• Using a strategic step-by-step approach to simplify complex calculations, such as applying the binomial probability formula.

By carefully following the steps, the seemingly intricate problem becomes more accessible. Each step builds upon the previous one to reach the overall solution. This iterative process of solving problems by breaking them down can be applied to various mathematical and real-world scenarios.

Combinatorics

Combinatorics is the field of mathematics concerned with counting, arranging, and understanding the structure of finite objects. It plays a key role in the solution of our problem through the concept of binomial distribution.

In binomial distribution, each trial is independent, and there are two possible outcomes - success or failure. In our exercise:

• The number of trials \( n \) equals 5 because there are five machines.
• Using combinatoric expressions, like \( \binom{n}{k} \), helps calculate the number of ways to choose \( k \) successes (machines out of service) from \( n \) trials.
• The formula \( \binom{5}{k} p^k q^{5-k} \) calculates the probability for each specific number \( k \) of machines being out of service, integrating probability theory with combinatorics.

Combinatorics doesn't just involve counting; it helps organize and make sense of probability by accounting for various possible outcomes by using systematic methods. It is critical in forming the basis for accurate probability calculations, as seen in this exercise where outcomes are meticulously counted and evaluated.