Distribution of the minimum of two independent exponential random variables

Can you please check if my reasoning for these questions is correct? I would really appreciate some help!

Q) Two types of cars A and B, arrive independently following an exponential distribution with rates of $k$ and $t$ per hour respectively.

a) Find the pdf of the time until you see the first car. What type of distribution do you see?

A: My reasoning is that since they are independent, time until the first car (considering both) can be found by the joint pdf, by multiplying the marginals (since they are independent). But will this again be an exponential distribution. Wouldn't it be something like $kt.e^{-t(k+t)}$. This is not the same as exponential distribution? (But I also have a hunch that we have to use T=min(A,B) here. That could be the time until the first car arrives.

b) Find the expected time until you see a total of $10$ cars? How many of them will be cars Type A? A: I am not sure how to proceed with this and the following question c) Suppose you saw $100$ cars today. Whats the probability you were at the gate for more than $8$ hours? Any help will be really appreciated. Thanks!

Let $X \sim \mathsf{Exp}(rate=\lambda_x = 3)$ and, independently, $Y \sim \mathsf{Exp}(rate=\lambda_y = 4).$ You seek the distribution of $V = \min(X,Y).$ As you speculate, $V$ also has an exponential distribution, but your formula for the PDF of $V$ is not quite right.
Here is an approach that will work. Because the minimum is less than $t$ precisely when both $X$ and $Y$ exceed $t,$ we have:
$$F_V(t) = P(V \le t) = P(X > t, Y > t) = P(X>t)P(Y>t) = (e^{-3t})(e^{-4t}) = e^{-7t},$$ for $t > 0.$
By differentiation, it follows that $V$ has the PDF of an exponential distribution with rate $\lambda_v = 7.$ And the PDF for your general case is one small step beyond that.