Question

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

To find the smallest integer greater than or equal to \(\frac{ln(0.15)}{ln(\frac{2}{3})}\), a calculator reveals this value to be approximately 3.47. Since \(n\) must be an integer, we round this up to 4. Therefore, the smallest value that \(n\) can take is 4.

Step-by-step solution

Identify Parameters and Establish the Complementary Rule

For a binomial distribution, the parameters include \(n\) (number of trials) and \(p\) (probability of success for each trial). As provided, \(p = \frac{1}{3}\). Our task is to find the smallest \(n\) for which \(P(X \geq 1) \geq 0.85\). This can be expressed as \(1 - P(X = 0) \geq 0.85\)

Expression for \(P(X = 0)\)

Using the formula of binomial distribution: \(P(X = k) = C(n, k) * (p)^k * (1-p)^{n-k}\) where \(C(n, k)\) is a combination function giving the number of ways to choose \(k\) successes from \(n\) trials. For \(P(X = 0)\) we get: \(P(X = 0) = (1 - p)^n\). Now, substitute \(1 - p\) into the expression, \(P(X = 0) = \left(\frac{2}{3}\right)^n\)

Find Minimum \(n\)

Since by the complementary rule \(1 - P(X = 0) \geq 0.85\), our equation becomes \(\left(\frac{2}{3}\right)^n \leq 0.15\). This is a simple inequality and we can solve it to find \(n\) by using logarithms. Taking the natural logarithm (ln) on both sides, we get: \(n \geq \frac{ln(0.15)}{ln(\frac{2}{3})}\).