\begin{tabular}{l} Andy has a dresser with two drawers, One drawer holds his socks, In it are 2 tan \\ pairs, \\ 4 black pairs, 3 blue pairs, and 6 white pairs of socks. The second drawer holds \\ his shirts, In it are 3 white shirts, 1 blue shirt, and 4 red shirts, Using this \\ information, find the following probabilities, \\ \begin{tabular}{l|l|l} 4. \( P \) (white socks, white shirt) & 5. P(black socks, red shirt) \end{tabular} \\ \( \begin{array}{ll}\text { 6. } P \text { (no white socks, no white shirt) } & \text { 7, P(black or blue socks, white or red } \\ \text { shirt) }\end{array} \) \\ \hline\end{tabular}

To find the probabilities for each of Andy's combinations of socks and shirts, we first need to determine the total number of socks and shirts. **Total Socks:** - Tan: 2 pairs - Black: 4 pairs - Blue: 3 pairs - White: 6 pairs Total Socks = 2 + 4 + 3 + 6 = 15 pairs **Total Shirts:** - White: 3 shirts - Blue: 1 shirt - Red: 4 shirts Total Shirts = 3 + 1 + 4 = 8 shirts **Calculating Probabilities:** 4. \( P \) (white socks, white shirt): This is the probability of selecting one white sock and one white shirt. \( P(\text{white socks}) = \frac{6}{15} \) and \( P(\text{white shirt}) = \frac{3}{8} \). Using the multiplication rule, \( P(\text{white socks, white shirt}) = P(\text{white socks}) \times P(\text{white shirt}) = \frac{6}{15} \times \frac{3}{8} = \frac{18}{120} = \frac{3}{20} \). 5. \( P(\text{black socks, red shirt}) \): \( P(\text{black socks}) = \frac{4}{15} \) and \( P(\text{red shirt}) = \frac{4}{8} = \frac{1}{2} \). So, \( P(\text{black socks, red shirt}) = \frac{4}{15} \times \frac{1}{2} = \frac{4}{30} = \frac{2}{15} \). 6. \( P(\text{no white socks, no white shirt}) \): To have no white socks, we consider tan, black, and blue socks. There are \( 15 - 6 = 9 \) non-white socks. Thus, \( P(\text{no white socks}) = \frac{9}{15} = \frac{3}{5} \). For shirts, there are \( 8 - 3 = 5 \) non-white shirts, giving \( P(\text{no white shirt}) = \frac{5}{8} \). Therefore, \( P(\text{no white socks, no white shirt}) = \frac{3}{5} \times \frac{5}{8} = \frac{15}{40} = \frac{3}{8} \). 7. \( P(\text{black or blue socks, white or red shirt}) \): For socks, \( P(\text{black or blue socks}) = P(\text{black socks}) + P(\text{blue socks}) = \frac{4}{15} + \frac{3}{15} = \frac{7}{15} \). For shirts, \( P(\text{white or red shirt}) = P(\text{white shirt}) + P(\text{red shirt}) = \frac{3}{8} + \frac{4}{8} = \frac{7}{8} \). Thus, \( P(\text{black or blue socks, white or red shirt}) = \frac{7}{15} \times \frac{7}{8} = \frac{49}{120} \). **Summary of Results:** 4. \( P \) (white socks, white shirt) = \( \frac{3}{20} \) 5. \( P \) (black socks, red shirt) = \( \frac{2}{15} \) 6. \( P \) (no white socks, no white shirt) = \( \frac{3}{8} \) 7. \( P \) (black or blue socks, white or red shirt) = \( \frac{49}{120} \)