A cashier counted the number of customers in their line every minute for 30 minutes. \( \begin{array}{llllll}\text { Find the probability that there were } \\ \text { exactly } 2 \text { customers in line. } \\ P(2)=\frac{9}{30} & P(2)=[?]\end{array} \)

To find the probability of having exactly 2 customers in line during the observation period, we can use the formula for probability: \[ P(X) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} \] In this case, the number of favorable outcomes for having exactly 2 customers counts as 9 (since it's been observed 9 times in 30 minutes). Therefore, \[ P(2) = \frac{9}{30} = 0.3 \] This means there's a 30% chance that at any given minute, the cashier will see exactly 2 customers in line! When it comes to probability distributions, the Poisson distribution might come in handy here. It's used to model the number of times an event occurs in a fixed interval of time. In this case, you could say that the occurrence of customers arriving is like random events, making it perfect for applying this distribution! Just remember, the Poisson formula is: \[ P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( \lambda \) is the average rate (in this case, \(\frac{9}{30}\) customers per minute), and \( k \) is the number of events you're interested in—in this case, \( k = 2 \).