A sample of 2 patients is selected from a group of 5 patients (2 shortsighted, 3 longsighted). Construct the probability distribution table for the number of shortsighted patients selected.

We use the combination formula $\binom{N}{n} = \frac{N!}{n!(N-n)!}$.

The number of ways to select $x$ shortsighted patients from $K$ shortsighted patients is $\binom{K}{x}$.
The number of ways to select $(n-x)$ longsighted patients from $(N-K)$ longsighted patients is $\binom{N-K}{n-x}$.
The probability $P(x)$ is given by:

• For $x=0$ (0 shortsighted, 2 longsighted): Number of ways to choose 0 shortsighted from 2: $\binom{2}{0} = 1$ Number of ways to choose 2 longsighted from 3: $\binom{3}{2} = \frac{3!}{2!1!} = 3$ Number of ways for $x=0$: $1 \times 3 = 3$ Probability $P(x=0) = \frac{3}{10}$

• For $x=1$ (1 shortsighted, 1 longsighted): Number of ways to choose 1 shortsighted from 2: $\binom{2}{1} = 2$ Number of ways to choose 1 longsighted from 3: $\binom{3}{1} = 3$ Number of ways for $x=1$: $2 \times 3 = 6$ Probability $P(x=1) = \frac{6}{10}$

• For $x=2$ (2 shortsighted, 0 longsighted): Number of ways to choose 2 shortsighted from 2: $\binom{2}{2} = 1$ Number of ways to choose 0 longsighted from 3: $\binom{3}{0} = 1$ Number of ways for $x=2$: $1 \times 1 = 1$ Probability $P(x=2) = \frac{1}{10}$

The sum of probabilities is $\frac{3}{10} + \frac{6}{10} + \frac{1}{10} = \frac{10}{10} = 1$, which confirms our calculations are correct.

The probability distribution table is: