Problem 46 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In a binomial distribution \(B\left(n, p=\frac{1}{4}\right)\), if the probability of at least one success is greater than equal to \(\frac{9}{10}\), then \(n\) is greater than (A) \(\frac{1}{\log _{10}^{4}-\log _{10}^{3}}\) (B) \(\frac{1}{\log _{10}{\underline{\phantom{xx}}}^{4}+\log _{10}{\underline{\phantom{xx}}}^{3}}\) (C) \(\frac{9}{\log _{10}{\underline{\phantom{xx}}}^{4}-\log _{10}^{3}}\) (D) \(\frac{4}{\log _{10}{\underline{\phantom{xx}}}^{4}-\log _{10}{\underline{\phantom{xx}}}^{3}}\)

Step-by-Step Solution

Answer

The correct option is (A).

Step 1: Understanding the Problem

We're given a binomial distribution with parameters \(n\) and \(p = \frac{1}{4}\). We need to find \(n\) such that the probability of at least one success is at least \(\frac{9}{10}\).

Step 2: Probability of At Least One Success

The probability of at least one success is given by the complement rule: 1 minus the probability of zero successes. For zero successes, the probability is \((1 - p)^n\), which here is \((\frac{3}{4})^n\). Thus, we have:\[ P(\text{at least one success}) = 1 - \left(\frac{3}{4}\right)^n \geq \frac{9}{10} \]

Step 3: Set Up the Inequality

Using the complement rule, our inequality becomes:\[ 1 - \left(\frac{3}{4}\right)^n \geq \frac{9}{10} \]This simplifies to:\[ \left(\frac{3}{4}\right)^n \leq \frac{1}{10} \]

Step 4: Apply Logarithms to Solve for n

Taking the natural logarithm on both sides, we get:\[ \log_{10}\left(\left(\frac{3}{4}\right)^n\right) \leq \log_{10}\left(\frac{1}{10}\right) \]Simplifying further:\[ n \log_{10}\left(\frac{3}{4}\right) \leq -1 \]

Step 5: Find the Threshold for n

Rearranging the inequality for \(n\):\[ n \geq \frac{-1}{\log_{10}\left(\frac{3}{4}\right)} \]Evaluating the logarithms using the property \[ \log_{10}\left(\frac{3}{4}\right) = \log_{10}3 - \log_{10}4 \], we have:\[ n \geq \frac{1}{\log_{10}4 - \log_{10}3} \]

Step 6: Comparison with Options

By comparing with the provided options, the closest and correct choice is **Option (A)**: \( n > \frac{1}{\log_{10}^4 - \log_{10}^3} \)].

Key Concepts

Probability of At Least One Success

In a binomial distribution scenario, determining the probability of at least one success can be easily done using the complement rule. Instead of calculating the probabilities of 1, 2, or up to n successes, we take a simpler route by calculating the probability of zero successes, and then subtracting it from 1. This process is much quicker, especially when dealing with many trials.
For instance, if the probability of an individual success is given by \( p \), then the probability of zero successes in \( n \) trials is \((1-p)^n\).
The probability of at least one success is therefore:

• \( P(\text{at least one success}) = 1 - (1-p)^n \)

In our exercise, with \( p = \frac{1}{4} \) and \( q = 1-\frac{1}{4} = \frac{3}{4} \), the task requires this result to be greater than or equal to \( \frac{9}{10} \). Substituting the values leads us to:

• \( 1 - \left(\frac{3}{4}\right)^n \geq \frac{9}{10} \)

This gives a simple way to find out the range for \( n \) by deducing an equation based on the complement rule.

Logarithmic Inequality

Solving inequalities involving exponents and powers often involves logarithms. They simplify the process of bringing powers down to values we can handle more straightforwardly. For example, if you come across an inequality like \( \left(\frac{3}{4}\right)^n \leq \frac{1}{10} \), applying logarithms is a practical next step.
By taking the logarithm on both sides, you transform your inequality to:

• \( \log_{10}\left(\left(\frac{3}{4}\right)^n\right) \leq \log_{10}\left(\frac{1}{10}\right) \)

Using the properties of logarithms, specifically that \( \log b^a = a \log b \), this becomes:

• \( n \log_{10}\left(\frac{3}{4}\right) \leq -1 \)

Rearranging things to solve for \( n \), you get:

• \( n \geq \frac{-1}{\log_{10} \left(\frac{3}{4}\right)} \)

This step allows us to conveniently address the problem of calculating the minimum number for which our condition (at least one success greater than or equal to \( \frac{9}{10} \)) holds true.

Complement Rule

One of the powerful tools in probability problems is the complement rule. When trying to find the probability of at least one event occurring, computing its complement—none of the events occurring—can sometimes be much simpler. According to this rule, \( P(A') \), the probability of the complement event (not A), plus \( P(A) \) equals 1.
Mathematically,

• \( P(A) = 1 - P(A') \)

For our binomial distribution problem, \( A \) is 'at least one success', and the complement, \( A' \), is 'zero successes'. By focusing on \( P(A') \), we use:

• \( P(\text{zero successes}) = (1-p)^n = \left(\frac{3}{4}\right)^n \)

Thus, calculating \( P(A) \) becomes much more manageable:

• \( P(\text{at least one success}) = 1 - \left(\frac{3}{4}\right)^n \)

This approach reduces potential computational complexity when \( n \) becomes large and straightforwardly guides us to consider suitable inequalities as shown in our exercise, where we set:

• \( 1 - \left(\frac{3}{4}\right)^n \geq \frac{9}{10} \)