1. Problem 13a: People immigrate according to a Poisson process with rate $\lambda = 2$ per day. Find the probability of 10 arrivals in 7 days. Step 1: The number of arrivals in 7 days is Poisson with mean $\mu = \lambda \times 7 = 2 \times 7 = 14$. Step 2: The probability of exactly $k=10$ arrivals is given by the Poisson pmf: $$ P(X=10) = \frac{e^{-14} 14^{10}}{10!} $$ Step 3: Calculate or approximate this probability using a calculator or software. 2. Problem 13b: Expectation of days until 20 arrivals. Step 1: The times between arrivals in a Poisson process are exponentially distributed. Step 2: The total time until $n=20$ arrivals follows a Gamma distribution with mean $\frac{n}{\lambda} = \frac{20}{2} = 10$ days. Step 3: Thus, expected time until 20 arrivals is $10$ days. 3. Problem 13b (battery life): Use CLT to approximate probability total lifetime of 25 batteries exceeds 1100 hours. Step 1: Given mean life $\mu=40$ hrs, sd $\sigma=20$ hrs, number $n=25$. Step 2: Total lifetime mean $= n\mu=1000$, variance $= n\sigma^2=25\times400=10000$, sd $=100$. Step 3: By CLT, total lifetime $S$ approximately normal. Step 4: Find $P(S > 1100) = P\left(Z > \frac{1100-1000}{100}\right) = P(Z > 1) \approx 0.1587$. 4. Problem 14a: Poisson arrivals with rate 3 per minute. Find probabilities on interarrival times. Interarrival time $T$ is exponentially distributed with rate $\lambda=3$ per min. (i) $P(T>1) = e^{-3 \times 1} = e^{-3}$. (ii) $P(1 < T < 2) = P(T > 1) - P(T > 2) = e^{-3} - e^{-6}$. (iii) $P(T \leq 4) = 1 - P(T > 4) = 1 - e^{-12}$. 5. Problem 14b: $X_1,\ldots,X_{10}$ independent Poisson(1). Use Markov inequality for $P(\sum X_i \geq 15)$. Step 1: Sum $S = \sum_{i=1}^{10} X_i$ is Poisson with mean $10$. Step 2: Markov inequality: $P(S \geq 15) \leq \frac{E[S]}{15} = \frac{10}{15} = \frac{2}{3}$. 6. Problem 15: Given TPM and initial distribution. (i) Compute $P(X_2=2)$. - Step 1: Find $P(X_1)$: $P(X_1) = P(0) \times P$. - Step 2: Find $P(X_2=2) = \sum_i P(X_1 = i) P_{i,2}$. (ii) $P(X_3=3,X_2=2,X_1=1,X_0=3)$ = $P(X_0=3)\times P_{3,1}\times P_{1,2}\times P_{2,3}$. (iii) $P^{(2)} = P \times P$ (matrix multiplication). 7. Problem 16: (a) Probability Capa wins all games Tuesday given she plays two games Tuesday. - Prob(win all) = $p^2$. (b) Expected games Wednesday: - Use Markov chain expected value:\ $E(G_{Wednesday}) = 1 \times P_{Tuesday=1, Wednesday=1} + 2 \times P_{Tuesday=1, Wednesday=2} + 1 \times P_{Tuesday=2, Wednesday=1} + 2 \times P_{Tuesday=2, Wednesday=2}$. Calculate using: If Tuesday=1 game, Wednesday distribution from first row: 0.2 and 0.8. If Tuesday=2 games, Wednesday distribution from second row: 0.4 and 0.6. Given Tuesday=2 games, $E = 1\times0.4 + 2\times0.6 = 0.4 + 1.2 = 1.6$. (c) Long run proportion of days winning all games: - Find stationary distribution $\pi$ solving $\pi P = \pi$. - Probability of winning all games on day is $\pi_1 \times p^1 + \pi_2 \times p^2$ (for days with 1 game or 2 games). 8. Problem Module 3, 13a (central limit theorem): - Given mean=1200, sd=250, sample size=60. - Probability average lifetime $>1250$. - Step 1: Sample mean sd = $\frac{250}{\sqrt{60}} \approx 32.26$. - Step 2: Compute $Z = \frac{1250-1200}{32.26} \approx 1.55$. - Step 3: Probability $P(\bar{X}>1250) = P(Z > 1.55) \approx 0.0606$. 9. Module 3 13b: Poisson process with rate 3 per minute, find probability exactly 4 arrive in 2 minutes. - Mean $\mu = 3 \times 2 = 6$. - $P(X=4) = e^{-6} \frac{6^4}{4!}$. 10. Module 3 14a: Weekly production mean=500, variance=100, find probability production between 400 and 600 using Chebyshev's inequality. - Interval $\pm 100$ from mean, $k=\frac{100}{\sqrt{100}}=10$. - Prob($|X-500| < 100$) $\geq 1 - \frac{1}{k^2} = 1 - \frac{1}{100} = 0.99$. 11. Module 3 14b(i-iii): Same as problem 4, exponential interarrival times with rate 3. 12. Module 4 5a: Markov chain states 0-4 TPM given. Find recurrent states. - Steps: States 0 and 1 communicate and form closed class. States 2 and 3 communicate and form closed class. State 4 can go to 0 or 1 or remain 4, so it communicates with 0,1. - Recurrent classes: {0,1,4} and {2,3}. 13. Module 4 5b: TPM given with initial distribution. (i) $P(X_2=3) = $ use total law: Calculate $P(X_1) = P(0) \times P$ Then $P(X_2=3) = \sum_i P(X_1=i) P_{i,3}$. (ii) Compute joint probability $P(X_3=2,X_2=3,X_1=3,X_0=2)$ = $P(X_0=2) \times P_{2,3} \times P_{3,3} \times P_{3,2}$.