Part a)

Let \(X\) be the number of customers arriving in a one-minute interval. Then \(X \sim Po(2.1),\) and

\(\Pr(X = 2) = \frac{\text{e}^{-2.1} 2.1^2}{2!}\) \(= 0.270.\)

Part b)

If \(Y\) denotes the number of customers arriving in a two-minute interval, then \(Y \sim Po(2 \times 2.1),\) and \(\Pr(Y > 3) = 1 - \Pr(Y \leq 3) = 0.000.\) Via R:

ppois(3, lambda=4.2, lower.tail=FALSE)

Part c)

For \(i = 1, 2, 3,\) let \(X_i\) be the number of customers arriving in minute \(i\text{.}\) These variables are independent, since counts in a Poisson process over non-overlapping intervals are independent; moreover \(X \sim Po(2.1).\) Hence

\(\begin{align*} \Pr(X_1 + X_2 + X_3 \geq 7 | X_1 = 2) \amp = \frac{\Pr(X_1 + X_2 + X_3 \geq 7, X_1 = 2)}{\Pr(X_1 = 2)} \\ \amp = \frac{\Pr(X_2 + X_3 \geq 5) \Pr(X_1 = 2)}{\Pr(X_1 = 2)} \\ \amp = \Pr(X_2 + X_3 \geq 5) \\ \amp = \Pr(Y \geq 5) \end{align*}\)

where \(Y \sim Po(4.2)\) as in part b. The result is \(0.000.\) Via R:

ppois(4, 4.2, lower.tail=FALSE)