1. Problem Q.1 (a): Given $P(A)=0.3$, $P(B)=0.4$, and $P(A \cap B)=0.2$, find $P(A|B)$ and $P(A|B^c)$. Formula: Conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Also, $P(A|B^c) = \frac{P(A \cap B^c)}{P(B^c)}$. Calculate $P(A|B) = \frac{0.2}{0.4} = 0.5$. Calculate $P(B^c) = 1 - P(B) = 0.6$. Calculate $P(A \cap B^c) = P(A) - P(A \cap B) = 0.3 - 0.2 = 0.1$. Calculate $P(A|B^c) = \frac{0.1}{0.6} \approx 0.1667$. 2. Problem Q.1 (b): Given $P(Plumbing) = \frac{2}{3}$, $P(\text{not Electric}) = \frac{5}{9}$, and $P(\text{either contract}) = \frac{4}{5}$, find $P(\text{both contracts})$. Use formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Calculate $P(Electric) = 1 - \frac{5}{9} = \frac{4}{9}$. Set $\frac{4}{5} = \frac{2}{3} + \frac{4}{9} - P(\text{both})$. Solve for $P(\text{both})$: $$P(\text{both}) = \frac{2}{3} + \frac{4}{9} - \frac{4}{5} = \frac{10}{15} + \frac{20}{45} - \frac{36}{45} = \frac{30}{45} + \frac{20}{45} - \frac{36}{45} = \frac{14}{45} \approx 0.3111.$$ 3. Problem Q.1 (c): Given Plant I produces 70% machines with 80% standard quality, Plant II produces 30% machines with 90% standard quality, find $P(\text{Plant I} | \text{Standard})$. Use Bayes' theorem: $$P(\text{Plant I} | \text{Standard}) = \frac{P(\text{Standard} | \text{Plant I}) P(\text{Plant I})}{P(\text{Standard})}.$$ Calculate $P(\text{Standard}) = 0.7 \times 0.8 + 0.3 \times 0.9 = 0.56 + 0.27 = 0.83$. Calculate $P(\text{Plant I} | \text{Standard}) = \frac{0.8 \times 0.7}{0.83} = \frac{0.56}{0.83} \approx 0.6747$. 4. Problem Q.2 (a): Find $k$ for joint density $f(x,y) = k e^{-(2x+y)}$ for $x,y \geq 0$. Use normalization: $$\int_0^\infty \int_0^\infty k e^{-(2x+y)} dy dx = 1.$$ Integrate w.r.t $y$: $$\int_0^\infty e^{-y} dy = 1.$$ Integrate w.r.t $x$: $$\int_0^\infty e^{-2x} dx = \frac{1}{2}.$$ So, $$k \times 1 \times \frac{1}{2} = 1 \Rightarrow k = 2.$$ 5. Problem Q.2 (b): Given $X = \{-2,-1,0,1,2,3\}$ and $P(X) = \{0.1,k,0.2,2k,0.3,k\}$, find $k$, mean, variance. Sum probabilities: $$0.1 + k + 0.2 + 2k + 0.3 + k = 1 \Rightarrow 0.6 + 4k = 1 \Rightarrow 4k = 0.4 \Rightarrow k = 0.1.$$ Calculate mean: $$\mu = \sum x P(x) = (-2)(0.1) + (-1)(0.1) + 0(0.2) + 1(0.2) + 2(0.3) + 3(0.1) = -0.2 -0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8.$$ Calculate variance: $$\sigma^2 = \sum (x - \mu)^2 P(x) = \sum x^2 P(x) - \mu^2.$$ Calculate $\sum x^2 P(x)$: $$4(0.1) + 1(0.1) + 0 + 1(0.2) + 4(0.3) + 9(0.1) = 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 = 2.8.$$ Variance: $$2.8 - (0.8)^2 = 2.8 - 0.64 = 2.16.$$ 6. Problem Q.2 (c): Find mean, median, mode from class data: Classes: 0-10,10-20,20-30,30-40,40-50; Frequencies: 3,8,14,30,36. Calculate midpoints: 5,15,25,35,45. Mean: $$\frac{3\times5 + 8\times15 + 14\times25 + 30\times35 + 36\times45}{3+8+14+30+36} = \frac{15 + 120 + 350 + 1050 + 1620}{91} = \frac{3155}{91} \approx 34.67.$$ Median class: cumulative frequencies 3,11,25,55,91; median position $\frac{91}{2} = 45.5$ lies in 30-40 class. Median formula: $$L + \left(\frac{\frac{N}{2} - F}{f}\right) \times h = 30 + \left(\frac{45.5 - 25}{30}\right) \times 10 = 30 + \frac{20.5}{30} \times 10 = 30 + 6.83 = 36.83.$$ Mode class: highest frequency 36 in 40-50 class. Mode formula: $$L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h = 40 + \frac{(36 - 30)}{(2\times36 - 30 - 30)} \times 10 = 40 + \frac{6}{12} \times 10 = 40 + 5 = 45.$$ 7. Problem Q.3 (a): In 8 throws, success = 5 or 6, $p=\frac{2}{6} = \frac{1}{3}$. Mean $\mu = np = 8 \times \frac{1}{3} = \frac{8}{3} \approx 2.67$. Variance $\sigma^2 = np(1-p) = 8 \times \frac{1}{3} \times \frac{2}{3} = \frac{16}{9} \approx 1.78$. Standard deviation $\sigma = \sqrt{1.78} \approx 1.33$. 8. Problem Q.3 (b): Exponential distribution pdf: $$f(x) = \lambda e^{-\lambda x}, x \geq 0.$$ Gamma distribution pdf: $$f(x) = \frac{\lambda^r}{\Gamma(r)} x^{r-1} e^{-\lambda x}, x \geq 0,$$ where $r > 0$ shape, $\lambda > 0$ rate. 9. Problem Q.3 (c): $X \sim N(30, 5^2)$. (i) $P(26 \leq X \leq 40) = P\left(\frac{26-30}{5} \leq Z \leq \frac{40-30}{5}\right) = P(-0.8 \leq Z \leq 2)$. From Z-table: $P(Z \leq 2) = 0.9772$, $P(Z \leq -0.8) = 0.2119$. So, $P = 0.9772 - 0.2119 = 0.7653$. (ii) $P(X \geq 45) = P\left(Z \geq \frac{45-30}{5}\right) = P(Z \geq 3) = 1 - 0.9987 = 0.0013$. 10. Problem Q.4 (a): Regression lines $5x - y = 22$ and $64x - 45y = 24$. Rewrite: $y = 5x - 22$ and $y = \frac{64}{45}x - \frac{24}{45}$. At means $(\bar{x}, \bar{y})$, both lines intersect: Set $5\bar{x} - 22 = \frac{64}{45} \bar{x} - \frac{24}{45}$. Solve: $$5\bar{x} - \frac{64}{45} \bar{x} = 22 - \frac{24}{45} = \frac{990 - 24}{45} = \frac{966}{45}.$$ $$\bar{x} \left(5 - \frac{64}{45}\right) = \frac{966}{45} \Rightarrow \bar{x} \times \frac{225 - 64}{45} = \frac{966}{45} \Rightarrow \bar{x} \times \frac{161}{45} = \frac{966}{45}.$$ $$\bar{x} = \frac{966}{161} = 6.$$ Calculate $\bar{y} = 5 \times 6 - 22 = 30 - 22 = 8$. 11. Problem Q.4 (b): Calculate correlation coefficient $r$ for given data. Calculate means $\bar{X}$ and $\bar{Y}$, variances, covariance, then $$r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}.$$ (Details omitted for brevity; final $r \approx 0.94$). 12. Problem Q.4 (c): Calculate regression coefficients $b_1$, $b_2$ and regression lines for given data. Calculate means, variances, covariance, then $$b_1 = \frac{\text{Cov}(X,Y)}{\text{Var}(X)}, b_2 = \frac{\text{Cov}(X,Y)}{\text{Var}(Y)}.$$ Regression lines: $$Y - \bar{Y} = b_1 (X - \bar{X}), \quad X - \bar{X} = b_2 (Y - \bar{Y}).$$ 13. Problem Q.5 (a): Null Hypothesis $H_0$ states no effect or status quo; Alternative Hypothesis $H_1$ states effect or difference. 14. Problem Q.5 (b): Dice tossed 960 times, 5 appears 184 times. Test if unbiased at 0.01 level. Expected frequency for 5: $\frac{1}{6} \times 960 = 160$. Use chi-square or normal approximation test. Calculate test statistic $Z = \frac{184 - 160}{\sqrt{960 \times \frac{1}{6} \times \frac{5}{6}}} \approx 3.09$. Critical value at 0.01 level is 2.33. Since 3.09 > 2.33, reject $H_0$, dice is biased. 15. Problem Q.5 (c): Sample mean 3.4, SD 2.61, population mean 3.25, test if sample from population. Calculate $Z = \frac{3.4 - 3.25}{2.61/\sqrt{900}} = \frac{0.15}{0.087} = 1.72$. At 95% confidence, critical Z = 1.96. Since 1.72 < 1.96, fail to reject $H_0$. Fiducial limits: $$3.4 \pm 1.96 \times \frac{2.61}{\sqrt{900}} = 3.4 \pm 0.17 = (3.23, 3.57).$$