1. **Problem statement:** We have a city where births occur at a rate of one birth every 12 minutes, and the time between births follows an exponential distribution. We need to find three things: i. The average number of births per year. ii. The probability that no births occur during 1 day. iii. The probability of issuing 50 birth certificates in 3 hours, given 40 were issued in the first 2 hours. 2. **Find the average number of births per year:** - Since there is one birth every 12 minutes, the rate \(\lambda\) is \(\frac{1}{12}\) births per minute. - Number of minutes in a year = \(365 \times 24 \times 60 = 525600\) minutes. - So, average births per year = \(\lambda \times \text{minutes per year} = \frac{1}{12} \times 525600 = 43800\). 3. **Calculate the probability that no births occur during 1 day:** - Number of minutes in 1 day = \(24 \times 60 = 1440\) minutes. - For an exponential distribution with rate \(\lambda = \frac{1}{12}\), the probability that no event occurs in time \(t\) is: $$ P(X > t) = e^{-\lambda t} $$ - Substitute \(\lambda = \frac{1}{12}\) and \(t = 1440\): $$ P(\text{no births in 1 day}) = e^{-\frac{1}{12} \times 1440} = e^{-120} $$ - Since \(e^{-120}\) is a very small number, the probability is effectively 0. 4. **Probability of issuing 50 birth certificates in 3 hours given 40 in first 2 hours:** - Birth process is Poisson since births are exponential inter-arrival times. - We want \(P(N(3) = 50 \mid N(2) = 40)\), number of births in 3 hours is 50 given 40 in first 2 hours. - The number of births in different intervals are independent increments. - The 3rd hour is from hour 2 to hour 3. - So the conditional probability is equivalent to \(P(N(3) - N(2) = 10)\). - The mean number of births per hour = \( \frac{60}{12} = 5 \) births/hour. - Number of births in the 3rd hour ~ Poisson with mean \(5\). - Probability: $$ P(N(1) = 10) = \frac{e^{-5} 5^{10}}{10!} $$ **Summary of answers:** - i. Average births/year = 43800 - ii. Probability of no births in 1 day = \(e^{-120}\) (very close to 0) - iii. Probability of 10 births in 1 hour = \(\frac{e^{-5} 5^{10}}{10!}\)