Chapter 7 - Section 7.1 - An Introduction to Discrete Probability - Exercises - Page 451: 10

5/663

There are $\binom{52}{5} = \frac{52!}{47!5!}$ equally likely poker hands. For the number of hands that contain 2D and 3Sp, there are $\binom{50}{3} = \frac{50!}{47!3!}$ choices. So the probability is $$\frac{50!}{47!3!} \frac{47!5!}{52!} = \frac{5(4)}{52(51)} = \frac{5}{663}$$