1. Problem: Given three boxes with capacitors of different values, find probabilities related to selecting capacitors. 2. For i) Probability of selecting a 0.01-μF capacitor given box 2 is selected: - Total capacitors in box 2 = 210 - Number of 0.01-μF capacitors in box 2 = 95 - Use conditional probability formula: $$P(0.01 \mid \text{box 2}) = \frac{\text{Number of 0.01 capacitors in box 2}}{\text{Total capacitors in box 2}} = \frac{95}{210}$$ 3. For ii) Probability that a selected 0.01-μF capacitor came from box 3: - Total 0.01-μF capacitors = 140 - Number of 0.01-μF capacitors in box 3 = 25 - Use Bayes' theorem: $$P(\text{box 3} \mid 0.01) = \frac{P(0.01 \mid \text{box 3}) P(\text{box 3})}{P(0.01)}$$ - Since each box is equally likely, $$P(\text{box 3}) = \frac{1}{3}$$ - Calculate $$P(0.01) = \sum_{i=1}^3 P(0.01 \mid \text{box } i) P(\text{box } i) = \frac{20}{145} \times \frac{1}{3} + \frac{95}{210} \times \frac{1}{3} + \frac{25}{245} \times \frac{1}{3}$$ - Simplify and compute: $$P(0.01) = \frac{20}{145} \times \frac{1}{3} + \frac{95}{210} \times \frac{1}{3} + \frac{25}{245} \times \frac{1}{3}$$ $$= \frac{20}{435} + \frac{95}{630} + \frac{25}{735}$$ - Find common denominator and sum: $$= \frac{20 \times 42}{18270} + \frac{95 \times 29}{18270} + \frac{25 \times 25}{18270} = \frac{840 + 2755 + 625}{18270} = \frac{4220}{18270}$$ - So, $$P(0.01) = \frac{4220}{18270}$$ - Now, $$P(0.01 \mid \text{box 3}) = \frac{25}{245}$$ - Therefore, $$P(\text{box 3} \mid 0.01) = \frac{\frac{25}{245} \times \frac{1}{3}}{\frac{4220}{18270}} = \frac{25}{245} \times \frac{1}{3} \times \frac{18270}{4220} = \frac{25 \times 18270}{245 \times 3 \times 4220}$$ - Simplify numerator and denominator: $$= \frac{456750}{3104700} = \frac{25}{170} \approx 0.1471$$ 4. b) i) Gaussian random variable X with mean $a_x=2$ and standard deviation $\sigma_x=2$. - Find $P\{X > 1.0\}$ and $P\{X \leq -1.0\}$. - Standardize using $Z = \frac{X - a_x}{\sigma_x}$. - For $P\{X > 1.0\}$: $$P\{X > 1.0\} = P\left\{Z > \frac{1.0 - 2}{2}\right\} = P\{Z > -0.5\} = 1 - P\{Z \leq -0.5\}$$ - Using symmetry, $P\{Z \leq -0.5\} = 1 - P\{Z \leq 0.5\}$. - From standard normal tables, $P\{Z \leq 0.5\} \approx 0.6915$. - So, $P\{X > 1.0\} = 1 - (1 - 0.6915) = 0.6915$. - For $P\{X \leq -1.0\}$: $$P\{X \leq -1.0\} = P\left\{Z \leq \frac{-1.0 - 2}{2}\right\} = P\{Z \leq -1.5\}$$ - From tables, $P\{Z \leq -1.5\} \approx 0.0668$. 5. ii) Find $b > 0$ so that $$f_X(x) = \begin{cases} \frac{e^x}{4}, & 0 \leq x \leq b \\ 0, & \text{elsewhere} \end{cases}$$ - For $f_X(x)$ to be a valid pdf, total area under curve must be 1: $$\int_0^b \frac{e^x}{4} dx = 1$$ - Compute integral: $$\frac{1}{4} \int_0^b e^x dx = \frac{1}{4} [e^x]_0^b = \frac{1}{4} (e^b - 1) = 1$$ - Solve for $b$: $$e^b - 1 = 4 \Rightarrow e^b = 5 \Rightarrow b = \ln 5$$ 6. iii) Conditional distribution definition and properties: - The conditional distribution of $Y$ given $X=x$ is the distribution of $Y$ when $X$ is fixed at $x$. - If joint pdf is $f_{X,Y}(x,y)$ and marginal pdf of $X$ is $f_X(x)$, then conditional pdf: $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)}$$ - Properties: 1. $f_{Y|X}(y|x) \geq 0$ for all $y$. 2. $$\int f_{Y|X}(y|x) dy = 1$$ for each fixed $x$. 3. Conditional expectation and variance can be defined using $f_{Y|X}(y|x)$. Final answers: - i) $$P(0.01 \mid \text{box 2}) = \frac{95}{210} \approx 0.4524$$ - ii) $$P(\text{box 3} \mid 0.01) \approx 0.1471$$ - b) i) $$P\{X > 1.0\} \approx 0.6915, \quad P\{X \leq -1.0\} \approx 0.0668$$ - ii) $$b = \ln 5 \approx 1.6094$$ - iii) Conditional distribution defined with properties as above.