Assignment: 1 whole sheet of paper Suppose three test kits are tested at random. Let \( D \) represent the defective test kit and let \( N \) represent the non-defective test kit. If we let \( X \) be the random variable for the number of defective test kits, construct the probability distribution of the random variable \( X \).

When dealing with three test kits, each kit can either be defective (D) or non-defective (N). The random variable \( X \), which represents the number of defective test kits, can take values from 0 to 3. To find the probability distribution, we'll assume that the probability of a defective test kit (D) is \( p \) and the probability of a non-defective test kit (N) is \( 1 - p \). 1. **X = 0 (No defective kits)**: - Probability: \( P(X=0) = (1-p)^3 \) 2. **X = 1 (One defective kit)**: - There are \( \binom{3}{1} = 3 \) ways to choose which of the three kits is defective. - Probability: \( P(X=1) = \binom{3}{1} p (1-p)^2 = 3p(1-p)^2 \) 3. **X = 2 (Two defective kits)**: - There are \( \binom{3}{2} = 3 \) ways to choose which two kits are defective. - Probability: \( P(X=2) = \binom{3}{2} p^2 (1-p) = 3p^2(1-p) \) 4. **X = 3 (All defective kits)**: - Probability: \( P(X=3) = p^3 \) Combining these, the probability distribution of \( X \) can be summarized as follows: - \( P(X=0) = (1-p)^3 \) - \( P(X=1) = 3p(1-p)^2 \) - \( P(X=2) = 3p^2(1-p) \) - \( P(X=3) = p^3 \) This distribution captures all possible scenarios of defective kits when sampling three test kits.