From a group of 8 people, 5 have skin rashes. If 5 people are selected randomly, what is the probability that exactly 3 have skin rashes and none have skin rashes?

This means selecting 3 people from the 5 with rashes AND $5-3=2$ people from the 3 without rashes.
Number of ways to select 3 with rashes: $\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$
Number of ways to select 2 without rashes: $\binom{3}{2} = \frac{3!}{2!1!} = 3$
The number of favorable ways is the product of these combinations:

This means selecting 0 people from the 5 with rashes AND $5-0=5$ people from the 3 without rashes.
Number of ways to select 0 with rashes: $\binom{5}{0} = 1$
Number of ways to select 5 without rashes: $\binom{3}{5} = 0$ (It is impossible to choose 5 people from a group of 3).
The number of favorable ways is the product of these combinations:

This means the number of people with rashes can be 0, 1, or 2.

• Case 1: 0 rashes (and 5 without rashes) Favorable ways = $\binom{5}{0} \times \binom{3}{5} = 1 \times 0 = 0$
• Case 2: 1 rash (and 4 without rashes) Number of ways to select 1 with rashes: $\binom{5}{1} = 5$ Number of ways to select 4 without rashes: $\binom{3}{4} = 0$ Favorable ways = $\binom{5}{1} \times \binom{3}{4} = 5 \times 0 = 0$
• Case 3: 2 rashes (and 3 without rashes) Number of ways to select 2 with rashes: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$ Number of ways to select 3 without rashes: $\binom{3}{3} = 1$ Favorable ways = $\binom{5}{2} \times \binom{3}{3} = 10 \times 1 = 10$ The total number of favorable ways for "at most 2 rashes" is the sum of these cases: