Problem 61 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

In a binomial distribution \(B\left(n, p=\frac{1}{4}\right)\), if the probability of at least one success is greater than or equal to \(\frac{9}{10}\), then \(n\) is greater than: [2009] (a) \(\frac{1}{\log _{10} 4+\log _{10} 3}\) (b) \(\frac{9}{\log _{10} 4-\log _{10} 3}\) (c) \(\frac{4}{\log _{10} 4-\log _{10} 3}\) (d) \(\frac{1}{\log _{10} 4-\log _{10} 3}\)

Step-by-Step Solution

Answer

The correct option is (d): \( \frac{1}{\log_{10}(4) - \log_{10}(3)} \).

Step 1: Define the Problem

We start with a binomial distribution \( B(n, p = \frac{1}{4}) \). The task is to find the condition under which the probability of at least one success is greater than or equal to \( \frac{9}{10} \).

Step 2: Express Probability Condition

The probability of at least one success is given by \( 1 - P(X = 0) \), where \( X \) is the number of successes. Thus, \( 1 - (1-p)^n \geq \frac{9}{10} \).

Step 3: Simplify the Expression

Substitute \( p = \frac{1}{4} \) into the expression: \( 1 - (\frac{3}{4})^n \geq \frac{9}{10} \). This implies \( (\frac{3}{4})^n \leq \frac{1}{10} \).

Step 4: Solve the Inequality with Logarithms

Take the logarithm of both sides: \(\log_{10}((\frac{3}{4})^n) \leq \log_{10}(\frac{1}{10}) \). Using log rules, it becomes \( n \cdot \log_{10}(\frac{3}{4}) \leq -1 \).

Step 5: Rearrange to Find n

Since \( \log_{10}(\frac{3}{4}) \) is negative, divide by it to get \( n \geq \frac{-1}{\log_{10}(\frac{3}{4})} \).

Step 6: Calculate \(log_{10}(3/4)) and Select Option

Use the properties of logarithms: \( \log_{10}(\frac{3}{4}) = \log_{10}(3) - \log_{10}(4) \). Thus, \( n \geq \frac{1}{\log_{10}(4) - \log_{10}(3)} \). Therefore, the correct answer is option (d): \( \frac{1}{\log_{10}(4) - \log_{10}(3)} \).

Key Concepts

Probability of success

In a binomial distribution, the probability of success \( p \) represents the likelihood that a single trial will result in success. In the given problem, each trial has a probability of success equal to \( \frac{1}{4} \). This means there's a 25% chance of success on each individual trial. We've used this probability to determine the likelihood of achieving at least one success in \( n \) trials.

The key focus here is understanding how the probability of at least one success is calculated. This is found by determining the probability that all trials result in failure and subtracting from one. Mathematically, it is shown as \( 1 - (1 - p)^n \).

In this formula:

• The term \( (1 - p)^n \) represents the probability that none of the \( n \) trials result in success.
• By subtracting this probability from 1, we find the probability of having at least one success, which needs to be greater than or equal to \( \frac{9}{10} \) as per the problem.

These calculations help us find the minimum number \( n \) of trials needed to reach the desired probability threshold.

Logarithmic inequalities

Logarithmic inequalities are critical for solving equations where variables are in an exponent, such as the inequality we encountered in this exercise. Once we established that the expression \( \left(\frac{3}{4}\right)^n \leq \frac{1}{10} \), taking the logarithm of both sides helped us extract \( n \) from the exponent.

Here's how we simplify the expression:

• Take logarithms on both sides: \( \log_{10}\left(\left(\frac{3}{4}\right)^n\right) \leq \log_{10}\left(\frac{1}{10}\right) \).
• Using the property of logarithms, \( \log_{10}\left(a^b\right) = b \cdot \log_{10}(a) \), it becomes \( n \cdot \log_{10}\left(\frac{3}{4}\right) \leq -1 \).

Finally, note that \( \log_{10}\left(\frac{3}{4}\right) \) is a negative number since \( \frac{3}{4} \lt 1 \). This requires us to handle the inequality direction carefully when dividing through by the logarithm to solve for \( n \), leading to \( n \geq \frac{-1}{\log_{10}\left(\frac{3}{4}\right)} \). This approach ensures proper handling and logic in inequality relationships.

Binomial theorem

The binomial theorem provides a formula for expanding expressions that are raised to a power. Although it wasn't directly applied in this problem, it aids in understanding binomial distributions. Such distributions reflect the number of successes in a series of yes/no experiments, each with the same probability of success.

For binomial distributions, the probability of obtaining exactly \( k \) successes in \( n \) trials with success probability \( p \) is given by:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]

• \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials.
• This formula captures how probabilities distribute across potential outcomes within a fixed number of trials.

Understanding this concept is foundational when working with binomial distributions and provides insight into how discrete probability unfolds in binomial contexts.

Discrete probability distribution

Discrete probability distributions describe situations where outcomes of a random variable can be counted. Binomial distribution is a perfect example of this, where each outcome represents a discrete count of successes. Each separate count has a specific probability, creating a full probability distribution for all possible outcomes.

Key characteristics of a discrete probability distribution include:

• Probabilities are assigned to individual outcomes.
• The sum of all probabilities is equal to 1.
• In our binomial context, each number of successful outcomes is a possible discrete outcome with its probability determined by the binomial formula.

Because of its structured nature, discrete probability distributions provide powerful models for determining the likelihood of various outcomes in trial-based events like those represented by binomial distributions. It's a highly systematic approach, allowing mathematicians and statisticians to predict the frequency of success types effectively.