Door to Door Sales

Solution

Answer: 45

A sale occurs with probability \(p = 0.05\) for each house visit. If the salesman visits \(n\) houses independently, the probability of making no sales is \[ (1 - 0.05)^n = 0.95^n. \] The probability of making at least one sale is therefore the complement: \[ P(\text{at least one sale}) = 1 - 0.95^n. \] We want this probability to be at least \(0.90\): \[ 1 - 0.95^n \ge 0.90. \] Rearranging, \[ 0.95^n \le 0.10. \] Taking natural logarithms of both sides, \[ n \ln(0.95) \le \ln(0.10). \] Solving for \(n\), \[ n \ge \frac{\ln(0.10)}{\ln(0.95)}. \] Substituting numerical values, \[ \ln(0.10) \approx -2.302585, \qquad \ln(0.95) \approx -0.051293. \] \[ n \ge \frac{-2.302585}{-0.051293} \approx 44.89. \] Since the number of visits must be a whole number, we round up: \[ n = 45. \] Checking: \[ P(\text{at least one sale}) = 1 - 0.95^{45} \approx 0.9006, \] which exceeds \(0.90\).