1. **Problem statement:** We have 77 female and 77 male applicants (total 154) for 5 positions. We want the probability of selecting exactly 3 females and 2 males when choosing 5 at random. 2. **Formula and rules:** The probability of selecting a specific combination is given by the hypergeometric distribution: $$P(X=k) = \frac{\binom{F}{k} \binom{M}{n-k}}{\binom{N}{n}}$$ where $F=77$ females, $M=77$ males, $N=154$ total, $n=5$ positions, and $k=3$ females selected. 3. **Calculate combinations:** - Number of ways to choose 3 females: $\binom{77}{3}$ - Number of ways to choose 2 males: $\binom{77}{2}$ - Total ways to choose any 5: $\binom{154}{5}$ 4. **Evaluate combinations:** $$\binom{77}{3} = \frac{77 \times 76 \times 75}{3 \times 2 \times 1} = 73150$$ $$\binom{77}{2} = \frac{77 \times 76}{2} = 2926$$ $$\binom{154}{5} = \frac{154 \times 153 \times 152 \times 151 \times 150}{5 \times 4 \times 3 \times 2 \times 1} = 6350135592$$ 5. **Calculate probability:** $$P = \frac{73150 \times 2926}{6350135592} = \frac{214176900}{6350135592} \approx 0.0337$$ 6. **Interpretation:** The probability of selecting exactly 3 females and 2 males is approximately 0.034 (rounded to three decimal places).