Problem 134
Question
In a binomial distribution \(\mathrm{B}\left(n, p=\frac{1}{4}\right)\), if the probability of at least one success is greater than or equal to \(\frac{9}{10}\), then \(\mathrm{n}\) is greater than \([2008]\) (A) \(\frac{1}{\log _{10}^{4}-\log _{10}^{3}}\) (B) \(\frac{1}{\log _{10}^{4}+\log _{10}^{3}}\) (C) \(\frac{9}{\log _{10}^{4}-\log _{10}^{3}}\) (D) \(\frac{4}{\log _{10}^{4}-\log _{10}^{3}}\)
Step-by-Step Solution
Verified Answer
The correct option is (A).
1Step 1: Understand the Problem
We have a binomial distribution with parameters \( n \) (number of trials) and \( p = \frac{1}{4} \) (probability of success). We need to determine the value of \( n \) such that the probability of at least one success is at least \( \frac{9}{10} \).
2Step 2: Express in Probability Terms
The probability of at least one success can be written as \( 1 - P(X = 0) \), where \( X \) is the binomial random variable. Here, \( P(X = 0) = (1-p)^n \). Therefore, the inequality \( 1 - (1-p)^n \geq \frac{9}{10} \) must hold.
3Step 3: Set up the Inequality
Simplifying the inequality: \[ 1 - (\frac{3}{4})^n \geq \frac{9}{10} \] Subtract 1 from both sides: \[ -(\frac{3}{4})^n \geq -\frac{1}{10} \] Multiply by \(-1\): \[ (\frac{3}{4})^n \leq \frac{1}{10} \]
4Step 4: Solve the Inequality
Take the logarithm of both sides:\[ n\log_{10}\left(\frac{3}{4}\right) \leq \log_{10}\left(\frac{1}{10}\right) \] Solving for \( n \): \[ n \geq \frac{\log_{10}\left(\frac{1}{10}\right)}{\log_{10}\left(\frac{3}{4}\right)} \] Noting that \( \log_{10}\left(\frac{1}{10}\right) = -1 \): \[ n \geq -\frac{1}{\log_{10}\left(\frac{3}{4}\right)} \]
5Step 5: Find the Value of n
Evaluating \( \log_{10}\left(\frac{3}{4}\right) = \log_{10}(3) - \log_{10}(4) \). Thus, \[ n \geq \frac{1}{\log_{10}(4) - \log_{10}(3)} \] This matches option (A).
6Step 6: Conclusion
The value of \( n \), ensuring the probability of at least one success is \( \geq \frac{9}{10} \), must be greater than or equal to option (A): \( \frac{1}{\log_{10}(4) - \log_{10}(3)} \).
Key Concepts
Probability TheoryBinomial CoefficientsInequality Solutions
Probability Theory
Probability theory is an essential corner of mathematics that helps us understand and quantify uncertainty. It's about predicting the likelihood of various outcomes. In the context of the binomial distribution, probability theory allows us to predict how likely it is to achieve a certain number of successes in a series of trials. Here, each trial has two possible outcomes: success or failure.
The binomial distribution is parameterized by two variables: the number of trials ( ), and the probability of success in a single trial ( p"). This distribution is crucial for solving problems where each trial is independent, and the probability of success remains constant through each trial.
Understanding probability in these terms lets us calculate outcomes like: "What are the odds of getting at least one success?" This involves calculating the probability of all other possible outcomes and subtracting it from 1, as seen in the exercise where we looked for the probability of at least one success in a series of trials.
The binomial distribution is parameterized by two variables: the number of trials ( ), and the probability of success in a single trial ( p"). This distribution is crucial for solving problems where each trial is independent, and the probability of success remains constant through each trial.
Understanding probability in these terms lets us calculate outcomes like: "What are the odds of getting at least one success?" This involves calculating the probability of all other possible outcomes and subtracting it from 1, as seen in the exercise where we looked for the probability of at least one success in a series of trials.
Binomial Coefficients
Binomial coefficients are instrumental in calculating probabilities in binomial distributions. They represent the number of ways to choose successes from a set of trials. The binomial coefficient is noted as \(\binom{n}{k}\), which reads as "n choose k", where \(n\) is the total number of trials and \(k\) is the number of successes.
Mathematically, binomial coefficients are expressed using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). It calculates how many different combinations of \(k\) successes can occur in \(n\) trials, discounting order. This concept generalizes the idea that some outcomes are more probable than others, based on the number of ways they can occur.
In many binomial probability problems, these coefficients are part of the probability calculations, especially where the direct outcome is fixed (like no successes in this exercise). Therefore, understanding and calculating binomial coefficients are vital to solving problems using binomial distribution effectively.
Mathematically, binomial coefficients are expressed using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). It calculates how many different combinations of \(k\) successes can occur in \(n\) trials, discounting order. This concept generalizes the idea that some outcomes are more probable than others, based on the number of ways they can occur.
In many binomial probability problems, these coefficients are part of the probability calculations, especially where the direct outcome is fixed (like no successes in this exercise). Therefore, understanding and calculating binomial coefficients are vital to solving problems using binomial distribution effectively.
Inequality Solutions
In mathematics, solving inequalities is a frequent task. It involves finding the range or specific values for variables that satisfy a given inequality. In probability problems using binomial distributions, understanding inequalities can help determine thresholds for desired outcomes.
In the example problem, we set up the inequality by rephrasing "at least one success" in a way that simplifies to an inequality \((1 - (\frac{3}{4})^n \geq \frac{9}{10})\). Solving this inequality required us to reorganize and manipulate the terms to isolate \(n\).
By applying logarithms, a technique that converts multiplication into addition (thus "undoing" exponential effects), we can simplify and solve the inequality to determine the minimum number of trials required. These steps illustrated how solving inequalities with logarithms clarifies equations involving exponentials, ultimately leading us to the algebraic solution of option (A) \(\frac{1}{\log_{10}(4) - \log_{10}(3)}\).
Understanding how to solve inequalities confidently ensures that you can determine thresholds effectively in probability and various other mathematical contexts.
In the example problem, we set up the inequality by rephrasing "at least one success" in a way that simplifies to an inequality \((1 - (\frac{3}{4})^n \geq \frac{9}{10})\). Solving this inequality required us to reorganize and manipulate the terms to isolate \(n\).
By applying logarithms, a technique that converts multiplication into addition (thus "undoing" exponential effects), we can simplify and solve the inequality to determine the minimum number of trials required. These steps illustrated how solving inequalities with logarithms clarifies equations involving exponentials, ultimately leading us to the algebraic solution of option (A) \(\frac{1}{\log_{10}(4) - \log_{10}(3)}\).
Understanding how to solve inequalities confidently ensures that you can determine thresholds effectively in probability and various other mathematical contexts.
Other exercises in this chapter
Problem 132
In the binomial expansion of \((a-b)^{n}, n \geq 5\), the sum of \(5^{\text {th }}\) and \(6^{\text {th }}\) terms is zero, then \(\frac{a}{b}\) equals [2007] (
View solution Problem 133
The sum of the series \({ }^{20} \mathrm{C}_{0}-{ }^{20} \mathrm{C}_{1}+{ }^{20} \mathrm{C}_{2}-{ }^{20} \mathrm{C}_{3}+\ldots-\ldots+{ }^{20} \mathrm{C}_{10}\)
View solution Problem 135
The remainder left out when \(8^{2 n}-(62)^{2 n+1}\) is divided by 9 is (A) 0 (B) 2 (C) 7 (D) 8
View solution Problem 136
The coefficient of \(x^{7}\) in the expansion of the expression \(\left(1-x-x^{2}+x^{3}\right)^{6}\) is (A) \(-132\) (B) \(-144\) (c) 132 (D) 144
View solution