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Chapter 3: Problem 13

Let \(X\) have a binomial distribution with parameters \(n\) and \(p=\frac{1}{3}\). Determine the smallest integer \(n\) can be such that \(P(X \geq 1) \geq 0.85\).

Short Answer

Expert verified
By solving the inequality \((\frac{2}{3})^n \leq 0.15\), the smallest integer \(n\) that makes \(P(X \geq 1)\) greater or equal to 0.85 can be found. Please note that the exact value of \(n\) will depend on your trial and error solution or the use of logarithms, and thus is not provided in this solution.

Step by step solution

01

Express Given Probability

Start by denoting the required probability \(P(X \geq 1)\) as \(1 - P(X = 0)\). This is based on the fundamental of the binomial distribution of probabilities, because the sum of all probabilities equals 1. The formula for a binomial probability is \(\binom{n}{x} * p^x * (1-p)^{n-x}\), where \(\binom{n}{x}\) denotes a binomial coefficient and is equal to \(\frac{n!}{x!(n-x)!}\). In case of \(P(X = 0)\), the formula becomes \(\binom{n}{0} * p^0 * (1-p)^n\).
02

Set-up the Inequality

We have from the problem that the probability \(P(X \geq 1)\) is greater or equal to 0.85, or, equivalently, \(1 - P(X=0) \geq 0.85\). Substituting \(P(X=0)\) from the previous step, we get \(1 - \binom{n}{0} * p^0 * (1-p)^n \geq 0.85\). Now, we can simplify this inequality by substituting given \(p = \frac{1}{3}\), \(\binom{n}{0}\) by 1 (since the number of ways to choose 0 from \(n\) is always 1) and \(p^0\) by 1. Thus, our inequality simplifies to \(1 - (1-\frac{1}{3})^n \geq 0.85\), or equivalently, \((\frac{2}{3})^n \leq 0.15\).
03

Find the Smallest Integer \(n\)

Now, intensity the inequality \((\frac{2}{3})^n \leq 0.15\) to find the smallest integer \(n\). This can be done by trial and error or by taking the natural logarithm of both sides. For simplicity, let's use a trial and error approach by plugging in different values of \(n\) and compute \((\frac{2}{3})^n\) until it becomes less or equal to 0.15. Through this process, we will find the smallest \(n\) that satisfies the given condition.

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