(c) \( n=8, p=0.36, X=0 \) \( P(X)=\square, 5 \)

To find the probability \( P(X) \) when \( n = 8 \), \( p = 0.36 \), and \( X = 0 \), we can use the binomial probability formula: \[ P(X) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials, - \( k \) is the number of successes, - \( p \) is the probability of success in a single trial, - \( X \) is the number of successes. Given: - \( n = 8 \), - \( p = 0.36 \), - \( X = 0 \). We need to find \( P(X) \) when \( X = 0 \), which means there are no successes. Substitute the given values into the formula: \[ P(X) = \binom{8}{0} (0.36)^0 (1-0.36)^{8-0} \] \[ P(X) = 1 \times 1 \times (0.64)^8 \] \[ P(X) = (0.64)^8 \] Now, we can calculate the value of \( P(X) \). Calculate the value by following steps: - step0: Calculate: \(0.64^{8}\) - step1: Convert the expressions: \(\left(\frac{16}{25}\right)^{8}\) - step2: Use the properties of exponents: \(\frac{16^{8}}{25^{8}}\) The probability \( P(X) \) when \( n = 8 \), \( p = 0.36 \), and \( X = 0 \) is approximately 0.028147.