1. **Problem statement:** A library has 20 fiction books and 30 non-fiction books, total 50 books. A student borrows 6 books randomly. We want the probability that less than 6 of the borrowed books are fiction. 2. **Understanding the problem:** "Less than 6 fiction books" means the student borrows 0, 1, 2, 3, 4, or 5 fiction books out of the 6. 3. **Formula used:** We use combinations to calculate probabilities in hypergeometric distribution: $$P(X = k) = \frac{\binom{20}{k} \binom{30}{6-k}}{\binom{50}{6}}$$ where $k$ is the number of fiction books borrowed. 4. **Calculate probability for each $k$ from 0 to 5 and sum:** $$P(X < 6) = \sum_{k=0}^5 P(X = k) = \sum_{k=0}^5 \frac{\binom{20}{k} \binom{30}{6-k}}{\binom{50}{6}}$$ 5. **Calculate total combinations:** $$\binom{50}{6} = 15,890,700$$ 6. **Calculate each term:** - $k=0$: $\binom{20}{0} \binom{30}{6} = 1 \times 593,775 = 593,775$ - $k=1$: $\binom{20}{1} \binom{30}{5} = 20 \times 142,506 = 2,850,120$ - $k=2$: $\binom{20}{2} \binom{30}{4} = 190 \times 27,405 = 5,206,950$ - $k=3$: $\binom{20}{3} \binom{30}{3} = 1,140 \times 4,060 = 4,628,400$ - $k=4$: $\binom{20}{4} \binom{30}{2} = 4,845 \times 435 = 2,106,075$ - $k=5$: $\binom{20}{5} \binom{30}{1} = 15,504 \times 30 = 465,120$ 7. **Sum numerator:** $$593,775 + 2,850,120 + 5,206,950 + 4,628,400 + 2,106,075 + 465,120 = 15,850,440$$ 8. **Calculate probability:** $$P(X < 6) = \frac{15,850,440}{15,890,700} \approx 0.9975$$ **Final answer:** The probability that less than 6 of the borrowed books are fiction is approximately **0.9975**.