A
(a) You're looking for
[tex]P(X\le 8) = \displaystyle \sum_{x=0}^8 P(X=x)[/tex]
where
[tex]P(X=x) = \begin{cases}\dfrac{\lambda^x e^{-\lambda}}{x!}&\text{if }x\in\{0,1,2,\ldots\}\\0&\text{otherwise}\end{cases}[/tex]
Customers arrive at a mean rate of 6 customers per 10 minutes, or equivalently 12 customers per 20 minutes, so
[tex]\lambda = \dfrac{12\,\rm customers}{20\,\rm min}\times(20\,\mathrm{min}) = 12\,\mathrm{customers}[/tex]
Then
[tex]\displaystyle P(X\le 8) = \sum_{x=0}^8 \frac{12^x e^{-12}}{x!} \approx \boxed{0.155}[/tex]
(b) Now you want
[tex]P(X\ge4) = 1 - P(X<4) = 1 - \displaystyle\sum_{x=0}^3 P(X=x)[/tex]
This time, we have
[tex]\lambda = \dfrac{6\,\rm customers}{10\,\rm min}\times(10\,\mathrm{min}) = 6\,\mathrm{customers}[/tex]
so that
[tex]P(X\ge4) = 1 - \displaystyle \sum_{x=0}^3 \frac{6^x e^{-6}}{x!} \approx \boxed{0.849}[/tex]
B
(a) In other words, you're asked to find the probability that more than 1 customer shows up in the same minute, or
[tex]P(X > 1) = 1 - P(X \le 1) = 1 - P(X=0) - P(X=1)[/tex]
with
[tex]\lambda = \dfrac{6\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 1\,\mathrm{customer}[/tex]
So we have
[tex]P(X > 1) = 1 - \dfrac{1^0 e^{-1}}{0!} - \dfrac{1^1 e^{-1}}{1!} \approx \boxed{0.264}[/tex]
C
(a) Similar to B, you're looking for
[tex]P(X \le 1) = P(X=0) + P(X=1)[/tex]
with
[tex]\lambda = \dfrac{12\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 2\,\mathrm{customers}[/tex]
so that
[tex]P(X\le1) = \dfrac{2^0e^{-2}}{0!} + \dfrac{2^1e^{-2}}{1!} \approx \boxed{0.406}[/tex]
