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.

This problem involves calculating probabilities for selecting patients without replacement, which is a scenario for a hypergeometric distribution. We need to form a probability distribution table for the random variable x, which represents the number of shortsighted patients selected. Here's the breakdown: Total patients: N = 5 Number of shortsighted patients: K = 2 Number of longsighted patients: N-K = 3 Number of patients selected: n = 2 Random variable x: number of shortsighted patients selected. The possible values for x (number of shortsighted patients selected) are 0, 1, or 2, since we are selecting 2 patients in total and there are only 2 shortsighted patients available. Step 1: Calculate the total number of ways to select 2 patients from the 5 available patients. We use the combination formula Nn = (N!)/(n!(N-n)!). Total combinations = 52 = (5!)/(2!(5-2)!) = (5!)/(2!3!) = (5 × 4 × 3 × 2 × 1)/((2 × 1)(3 × 2 × 1)) = (20)/(2) = 10 Step 2: Calculate the number of ways to select x shortsighted patients and (2-x) longsighted patients for each possible value of x. The number of ways to select x shortsighted patients from K shortsighted patients is Kx. The number of ways to select (n-x) longsighted patients from (N-K) longsighted patients is N-Kn-x. The probability P(x) is given by: P(x) = (K)/(x) N-Kn-xNn For x=0 (0 shortsighted, 2 longsighted): Number of ways to choose 0 shortsighted from 2: 20 = 1 Number of ways to choose 2 longsighted from 3: 32 = (3!)/(2!1!) = 3 Number of ways for x=0: 1 × 3 = 3 Probability P(x=0) = (3)/(10) For x=1 (1 shortsighted, 1 longsighted): Number of ways to choose 1 shortsighted from 2: 21 = 2 Number of ways to choose 1 longsighted from 3: 31 = 3 Number of ways for x=1: 2 × 3 = 6 Probability P(x=1) = (6)/(10) For x=2 (2 shortsighted, 0 longsighted): Number of ways to choose 2 shortsighted from 2: 22 = 1 Number of ways to choose 0 longsighted from 3: 30 = 1 Number of ways for x=2: 1 × 1 = 1 Probability P(x=2) = (1)/(10) Step 3: Form the probability distribution table. The sum of probabilities is (3)/(10) + (6)/(10) + (1)/(10) = (10)/(10) = 1, which confirms our calculations are correct. |c|c| x & P(x) \\ 0 & (3)/(10) \\ 1 & (6)/(10) \\ 2 & (1)/(10) \\ The probability distribution table is: |c|c| x & P(x) \\ 0 & (3)/(10) \\ 1 & (6)/(10) \\ 2 & (1)/(10) \\ That's 2 down. 3 left today — send the next one.