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 $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Step 1: Calculate the probability of making $k$ sales for $k=0, 1, 2, 3, 4, 5$. $$ P(X=0) = \binom{5}{0} (0.30)^0 (0.70)^5 = 1 \times 1 \times 0.16807 = 0.16807 $$ $$ P(X=1) = \binom{5}{1} (0.30)^1 (0.70)^4 = 5 \times 0.30 \times 0.2401 = 0.36015 $$ $$ P(X=2) = \binom{5}{2} (0.30)^2 (0.70)^3 = 10 \times 0.09 \times 0.343 = 0.30870 $$ $$ P(X=3) = \binom{5}{3} (0.30)^3 (0.70)^2 = 10 \times 0.027 \times 0.49 = 0.13230 $$ $$ P(X=4) = \binom{5}{4} (0.30)^4 (0.70)^1 = 5 \times 0.0081 \times 0.70 = 0.02835 $$ $$ P(X=5) = \binom{5}{5} (0.30)^5 (0.70)^0 = 1 \times 0.00243 \times 1 = 0.00243 $$ A) Probability of at most 3 sales, $P(X \le 3)$. This is the sum of probabilities for $X=0, 1, 2, 3$. $$ P(X \le 3) = P(X=0) + P