Exponential distribution probabilities and memoryless property

Question

Question:

You are at a restaurant with two other friends waiting for food. You and both of your friends have each ordered one plate. Assume the number of plates brought to the table in an hour follows a Poisson distribution with parameter $λ=6$.

1. What is the probability that one of the plates arrives within 5 minutes?

2. There's a rule that nobody can eat until everyone receives their food. What is the probability that it will take longer than 15 minutes to eat (longer than 15 minutes for all three plates to arrive?)

3. You've already waited 10 minutes and the first plate hasn't arrived. What is the probability you'll have to wait ten additional minutes for it? How many additional minutes should you expect to wait for the plate?

My attempt:

With a Poisson distribution of $λ=6$ plates per hour, this is equivalent to $λ=\frac{6}{60}=\frac{1}{10}$ plates per minute.

1. The probability density function of an exponential distribution is $f(x)=λe^{-λx}$. So $P(X≤5) = \int_{0}^5 λe^{-λx}$ where again, $λ=\frac{1}{10}$. Which evaluates to $\int_{0}^5 \frac{1}{10}e^{-\frac{1}{10}x} = 1-e^{\frac{-1}{2}}=.3934$.

2. I believe I should be using the gamma distribution here. The gamma distribution can be written as $f(x)=\frac{1}{Γ(k)θ^k}x^{k-1}e^{-\frac{x}{θ}}$. Where we have the parameters $k=3$ and $θ=10$. Of course since we are looking at $P(X≥15)$ we define the bounds to be from 15 minutes to infinity minutes. We would then simply solve to find $\int_{15}^\infty \frac{1}{Γ(k)θ^k}x^{k-1}e^{-\frac{x}{θ}}$ and after plugging in the parameters we get, $\int_{15}^\infty \frac{1}{Γ(3)10^3}x^{3-1}e^{-\frac{x}{10}}=.80885$.

3. I think this component should utilize the "memoryless property" of exponential distributions. We are trying to find $P$(wait ≥ 20 minutes | wait ≥ 10 minutes). Let X be a random variable that represents the number of minutes we wait. Then: $\frac{P(X≥20 \cap X≥10)}{P(X≥10)}=\frac{P(X≥20)}{P(X≥10)}=\frac{1-P(X≤20)}{1-P(X≤10)}=\frac{e^{-20λ}}{e^{-10λ}}=e^{-10λ}=P(X≥10)$. We can evaluate to find this expression given that we know $λ$ as well, giving us $e^{-10\frac{1}{10}}=e^{-1}$.

To find the number of additional minutes we expect to wait: we know that $E[X]=\frac{1}{λ}$ and we know that $λ=\frac{1}{10}$. So the expected number of additional minutes we will wait is $E[X]=\frac{1}{\frac{1}{10}}=10$.