Question: A college bookstore receives a shipment of 24 laptops for students. During an initial check, it is determined that 4 of the laptops are defective. Before placing the laptops on sale, an employee randomly selects 5 laptops for inspection.
What is the probability that exactly 2 of the 5 selected laptops are defective?
Round your answer to three decimal places.
1. **State the problem:** We want to find the probability that exactly 2 out of 5 randomly selected laptops are defective, given there are 4 defective laptops out of 24 total.
2. **Formula used:** This is a hypergeometric probability problem. The formula is:
$$P(X = k) = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}}$$
where:
- $N = 24$ (total laptops),
- $D = 4$ (defective laptops),
- $n = 5$ (laptops selected),
- $k = 2$ (defective laptops selected).
3. **Calculate combinations:**
- $\binom{4}{2} = \frac{4!}{2! (4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$
- $\binom{20}{3} = \frac{20!}{3! (20-3)!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$
- $\binom{24}{5} = \frac{24!}{5! (24-5)!} = \frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1} = 42504$
4. **Plug values into formula:**
$$P(X=2) = \frac{6 \times 1140}{42504} = \frac{6840}{42504}$$
5. **Simplify fraction:**
$$\frac{6840}{42504} = \frac{\cancel{6840}}{\cancel{42504}}$$
(We simplify by dividing numerator and denominator by 12):
$$\frac{6840 \div 12}{42504 \div 12} = \frac{570}{3542}$$
6. **Calculate decimal:**
$$P(X=2) \approx \frac{570}{3542} \approx 0.161$$
**Final answer:** The probability that exactly 2 of the 5 selected laptops are defective is approximately **0.161**.