1. **Problem Statement:** We have 3 cell phones tested at random, and $X$ is the random variable representing the number of defective cell phones among the 3 tested. 2. **Assumptions:** We assume each cell phone can be either defective or not defective, and the probability of a phone being defective is $p$. The number of defective phones $X$ follows a binomial distribution because each test is independent and has two outcomes (defective or not). 3. **Formula:** The probability mass function (PMF) of a binomial distribution is: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where: - $n = 3$ (number of trials) - $k = 0,1,2,3$ (number of defective phones) - $p$ = probability a phone is defective 4. **Constructing the Probability Distribution:** - $P(X=0) = \binom{3}{0} p^0 (1-p)^3 = 1 \times 1 \times (1-p)^3 = (1-p)^3$ - $P(X=1) = \binom{3}{1} p^1 (1-p)^2 = 3p(1-p)^2$ - $P(X=2) = \binom{3}{2} p^2 (1-p)^1 = 3p^2(1-p)$ - $P(X=3) = \binom{3}{3} p^3 (1-p)^0 = p^3$ 5. **Explanation:** Each term represents the probability of exactly $k$ defective phones out of 3. The binomial coefficient counts the ways to choose which phones are defective. 6. **Final Probability Distribution:** $$\begin{cases} P(X=0) = (1-p)^3 \\ P(X=1) = 3p(1-p)^2 \\ P(X=2) = 3p^2(1-p) \\ P(X=3) = p^3 \end{cases}$$ This distribution fully describes the probabilities for the number of defective phones tested.