1. **Problem Statement:** We are testing 3 cellphones and want to find the probability distribution of the random variable $x$, which represents the number of defective cellphones. Let $D$ be defective and $N$ be non-defective. 2. **Understanding the Random Variable:** The random variable $x$ can take values $0, 1, 2,$ or $3$ because there can be from zero to all three cellphones defective. 3. **Assumptions:** We assume each cellphone is defective independently with some probability $p$. Since $p$ is not given, we will express probabilities in terms of $p$. 4. **Probability Distribution Formula:** The number of defective cellphones $x$ follows a binomial distribution: $$P(x=k) = \binom{3}{k} p^k (1-p)^{3-k}$$ where $k = 0,1,2,3$. 5. **Calculating Each Probability:** - $P(x=0) = \binom{3}{0} p^0 (1-p)^3 = (1-p)^3$ - $P(x=1) = \binom{3}{1} p^1 (1-p)^2 = 3p(1-p)^2$ - $P(x=2) = \binom{3}{2} p^2 (1-p)^1 = 3p^2(1-p)$ - $P(x=3) = \binom{3}{3} p^3 (1-p)^0 = p^3$ 6. **Summary:** The probability distribution of $x$ is: $$\begin{aligned} P(x=0) &= (1-p)^3 \\ P(x=1) &= 3p(1-p)^2 \\ P(x=2) &= 3p^2(1-p) \\ P(x=3) &= p^3 \end{aligned}$$ This distribution gives the probabilities of having 0, 1, 2, or 3 defective cellphones out of 3 tested, depending on the defect probability $p$.