This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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This problem involves a binomial distribution, but the number of sales contacts (n) is not explicitly stated. We will assume a reasonable number of contacts to solve the problem. Assumption: We assume the sales manager makes n=5 sales contacts in a particular day. If the actual number of contacts is different, the probabilities will change accordingly. Given: • Probability of a sale (p) = 0.30 • Probability of no sale (1-p) = 1 - 0.30 = 0.70 • Number of contacts (n) = 5 (assumed) Let X be the number of sales made. The probability mass function for a binomial distribution is P(X=k) = nk p^k (1-p)^n-k. Step 1: Calculate the probability of making k sales for k=0, 1, 2, 3, 4, 5. P(X=0) = 50 (0.30)^0 (0.70)^5 = 1 × 1 × 0.16807 = 0.16807 P(X=1) = 51 (0.30)^1 (0.70)^4 = 5 × 0.30 × 0.2401 = 0.36015 P(X=2) = 52 (0.30)^2 (0.70)^3 = 10 × 0.09 × 0.343 = 0.30870 P(X=3) = 53 (0.30)^3 (0.70)^2 = 10 × 0.027 × 0.49 = 0.13230 P(X=4) = 54 (0.30)^4 (0.70)^1 = 5 × 0.0081 × 0.70 = 0.02835 P(X=5) = 55 (0.30)^5 (0.70)^0 = 1 × 0.00243 × 1 = 0.00243 A) Probability of at most 3 sales, P(X 3). This is the sum of probabilities for X=0, 1, 2, 3. P(X 3) = P(X=0) + P