Exponential distribution of runners

Question

Neil and Patrick go for runs whose lengths are independent and identically distributed exponential random varaibles with $\mu =35$ minutes.

1) What is the probability that the first to finish takes atleast 35 minutes to finish running.

My logic: Note $Pr(min(A,B) \geq z) = Pr(A \geq z) \cdot Pr(B \geq z)$, so since they are identically distributed exponential random variables, it should just be $2 \cdot Pr(A \geq z)$. Where $Pr(A \geq z)$ is calculated by $P = \lambda e^{-\lambda x}$, where $\lambda = 1/35, x = 35$ so $P \approx .0105$, so the answer is $2*P = .02102$?

2) What is the probability that Neil's run takes atleast 35 minutes?

My logic: Should just be P as above? So $.0105$?

I feel like I have done this wrong. Can I have confirmation or tips please, would be greatly appreciated.

Answer 1

Note $Pr(min(A,B) \geq z) = Pr(A \geq z) \cdot Pr(B \geq z)$, so since they are identically distributed exponential random variables, it should just be $2 \cdot Pr(A \geq z)$.

No.

...since they are identically distributed exponential random variables and independent, it should just be $\Pr(A \geq z)^2$.

Where $Pr(A \geq z)$ is calculated by $P = \lambda e^{-\lambda x}$

No.

...where $\Pr(A \geq z)$ is by definition $p=\displaystyle\int_z^\infty\lambda\mathrm e^{-\lambda t}\mathrm dt=\mathrm e^{-\lambda z}$.

2) What is the probability that Neil's run takes atleast 35 minutes?

My logic: Should just be P as above?

Yes, if $P$ is the correct $p=\mathrm e^{-\lambda z}$ computed above, that is, $p=1/\mathrm e$.