You are a new business owner and ordered 200 promotional smart phone cases to give out to clients. The seller has said that 8 of the cases are defective. You unpack the package and inspect 15 of the cases. 1. What is the probability that 10 are defective? 2. What is the probability that all 15 are defective?

We start by noting that the total population has \( N = 200 \) phone cases, of which the number of defective cases is \( K = 8 \). You randomly select \( n = 15 \) of these cases. The probability of drawing exactly \( k \) defective cases in the sample is given by the hypergeometric probability formula: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}, \] where - \( \binom{a}{b} \) is the binomial coefficient, - \( K = 8 \) is the total number of defective cases, - \( N-K = 200-8 = 192 \) is the number of non-defective cases, - \( n = 15 \) is the sample size, - \( k \) is the number of defective cases in the sample. --- 1. For the probability that exactly 10 cases are defective, set \( k = 10 \): \[ P(X = 10) = \frac{\binom{8}{10} \binom{192}{15 - 10}}{\binom{200}{15}}. \] Notice that \(\binom{8}{10} = 0\) because you cannot choose 10 items from a set of 8. Thus, \[ P(X = 10) = 0. \] --- 2. For the probability that all 15 cases are defective, set \( k = 15 \): \[ P(X = 15) = \frac{\binom{8}{15} \binom{192}{15 - 15}}{\binom{200}{15}}. \] Again, \(\binom{8}{15} = 0\) since 15 is greater than 8. Therefore, \[ P(X = 15) = 0. \] --- Thus, the probability that 10 are defective is \( 0 \) and the probability that all 15 are defective is also \( 0 \).