As mentioned in the comments, the answer depends very much on the model used to describe the passage times of the buses. The deterministic situation where the passage times of buses of type $k$ are $s_k+m_k\mathbb N$ for some initial passage time $s_k$ in $(0,m_k)$ is too unwieldy to be dealt with in full generality hence we now study two types of assumptions.
(1) Fully random passage times
Here the passage times of buses of type $k$ are a Poisson process of intensity $1/m_k$ and the passage times of buses of different types are independent. Then, starting at time $t_0$, the next bus of type $k$ arrives after a random time exponential with mean $m_k$ hence the waiting time $T$ is such that
$$
\mathbb P(T\gt t)=\prod_k\mathbb P(\text{no bus of type}\ k\ \text{in}\ (t_0,t_0+t))=\prod_k\mathrm e^{-t/m_k}=\mathrm e^{-t/m},
$$
where
$$
\frac1m=\sum_k\frac1{m_k}.
$$
In particular, $T$ is exponentially distributed with parameter $1/m$, hence
$$
\mathbb E(T)=m.
$$
The case $m_1=m_2=\cdots=m_n$ yields
$$
\mathbb E(T)=\frac{m_1}{n}.
$$
(2) Fully periodic passage times with random uniform initializations
Here, buses of type $k$ pass at times in $S_k+m_k\mathbb N$ where $S_k$ is uniform on $(0,m_k)$ and the random variables $(S_k)$ are independent. Now, starting at time $t_0$, the next bus of type $k$ arrives after time $t_0+t$ if $t\leqslant m_k$ and if $S_k$ is not in a subinterval of $(0,m_k)$ of lenth $t/m_k$. Thus,
$$
\mathbb P(T\gt t)=\prod_k\left(1-\frac{t}{m_k}\right),\qquad t\leqslant \bar m=\min\limits_km_k.
$$
A consequence is that
$$
\mathbb E(T)=\int_0^{+\infty}\mathbb P(T\gt t)\,\mathrm dt=\int_0^{\bar m}\prod_k\left(1-\frac{t}{m_k}\right)\,\mathrm dt.
$$
Expanding the product yields
$$
\mathbb E(T)=\sum_{i\geqslant0}(-1)^i\bar m^{i+1}\frac1{i+1}\sum_{|K|=i}\frac1{m_K},
$$
where, for every subset $K$,
$$
m_K=\prod_{k\in K}m_k.
$$
For example, time intervals $m_1$, $m_2$, $m_3$ with minimum $m_1$ yield
$$
\mathbb E(T)=m_1-\frac{m_1^2}2\left(\frac1{m_1}+\frac1{m_2}+\frac1{m_3}\right)+\frac{m_1^3}{3}\left(\frac1{m_1m_2}+\frac1{m_2m_3}+\frac1{m_3m_1}\right)-\frac{m_1^4}{4m_1m_2m_3},
$$
which can be simplified a little bit (but not much) into
$$
\mathbb E(T)=\frac{m_1}2-\frac{m_1^2}{6m_2}-\frac{m_1^2}{6m_3}+\frac{m_1^3}{12m_2m_3}.
$$
The case $m_1=m_2=\cdots=m_n$ yields
$$
\mathbb E(T)=\frac{m_1}{n+1}.
$$
Edit: The wording has changed in a way that makes the problem entirely different. We keep the answer to the original problem below, and show how to solve the new problem at the end of this post.
Solution to old problem: Using an exponential distribution model seems unreasonable, but you are probably expected to do so. Then use the fact that the exponential distribution is memoryless. The waiting time, from the time you arrive, until the first bus has exponential distribution with mean $15$ minutes.
Calculation: Our exponential has mean $15$, and therefore density function $\frac{1}{15}e^{-x/15}$ (for $x\gt 0$). Thus if $X$ is the waiting time, then
$$\Pr(X\lt 10)=\int_0^{10} \frac{1}{15}e^{-x/15}=1-e^{-10/15}.$$
The new problem: We assume that we arrive at the stop at a time uniformly distributed between 7 and 7:30 (unreasonable!). Then our waiting time is less than $10$ minutes if we arrive between 7:05 and 7:15, or between 7:20 and 7:30. That has total length $20$ minutes, so the probability our arrival time is in this region is $\frac{20}{30}$.
Best Answer
It's useful to split this into two cases as a function of the arrival time. Let event $A_0=$ arrival at bus stop before 40 minutes past the hour (that is, $\geq 0$ but $<40$ minutes), and $A_1=$ arrival at bus stop $\geq 40$ minutes past the hour. These events have probabilitites $$ P(A_0) = \frac{40}{60} = \frac{2}{3}; \qquad P(A_1) = \frac{1}{3} $$
Now we can consider the waiting time for a bus. Let $S$ be the waiting time in minutes. Now the CDF is $$ \begin{split} F_S(S) = P(S<s) &= P(S<s|A_0)P(A_0) + P(S<s | A_1)P(A_1) \\ \end{split} $$ The term $P(S=s|A_0)$ is evenly distributed between $0$ and $40$, and the term $P(S=s | A_1)$ is evenly distributed between $0$ and $20$, and therefore $$ \begin{split} F_S(s) = P(S<s) &= P(S<s|A_0)P(A_0) + P(S<s | A_1)P(A_1) \\ &= \text{Uniform}[0,40] \frac{2s}{3} + \text{Uniform}[0,20]\frac{1s}{3} \end{split} $$ Therefore, the PDF is (just the derivative of $F_S$) $$ f_S(s) = \left\{ \begin{array}{ccc} \frac{1}{30},& 0 \leq s \leq 20 \\ \frac{1}{60},& 20 \leq s \leq 40 \\ 0 , & \text{otherwise} \end{array} \right. $$ The mean value is $$ \mathbb{E}[S] = \int_0^{40}f_S(s') s' \, ds' = \int_0^{20} \frac{s'}{30} \, ds' + \int_{20}^{40} \frac{s'}{60} \, ds' = \frac{20}{3} + 10 \approx 16.6667 $$