1. **Problem Statement:** We are given that 35% of employees who apply for promotion are successful. Six employees apply, and we want to find the probability that at most four employees are successful. 2. **Formula Used:** This is a binomial probability problem. The probability of exactly $k$ successes in $n$ trials is given by: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=6$, $p=0.35$, and $k$ is the number of successful employees. 3. **At most four successes** means $X \leq 4$, so we calculate: $$P(X \leq 4) = \sum_{k=0}^4 P(X = k)$$ 4. **Calculate each term:** - $P(X=0) = \binom{6}{0} 0.35^0 (0.65)^6 = 1 \times 1 \times 0.65^6 = 0.0759$ - $P(X=1) = \binom{6}{1} 0.35^1 (0.65)^5 = 6 \times 0.35 \times 0.65^5 = 0.2433$ - $P(X=2) = \binom{6}{2} 0.35^2 (0.65)^4 = 15 \times 0.1225 \times 0.65^4 = 0.3241$ - $P(X=3) = \binom{6}{3} 0.35^3 (0.65)^3 = 20 \times 0.0429 \times 0.65^3 = 0.2363$ - $P(X=4) = \binom{6}{4} 0.35^4 (0.65)^2 = 15 \times 0.0150 \times 0.65^2 = 0.0933$ 5. **Sum the probabilities:** $$P(X \leq 4) = 0.0759 + 0.2433 + 0.3241 + 0.2363 + 0.0933 = 0.9729$$ 6. **Interpretation:** The probability that at most four employees are successful is approximately 0.973, which is closest to option D (0.978). **Final answer:** $\boxed{0.978}$