Let $p$ be the probability of success (here $p = 0.3$), and $q = 1 - p$, the probability of failure (here $q = 0.7$). $X = k$ if there are $k - 1$ trials with exactly $2$ successes and $k - 3$ failures, and the $k$th trial is success.
The probability of exactly $2$ successes in a sequence of $k - 1$ trials is $^{k-1}\text{C}_2\, p^2 q^{k - 3}$ (by using a binomial distribution with $n = k - 1$). Multiplying this with $p$, the probability of success in the $k$th trial, we obtain
$$P(X = k) = ^{k-1}\text{C}_2\, p^3 q^{k - 3},\ k = 3, 4, \ldots$$
$\begin{align}
E[X] & = \sum_{k = 3}^{\infty} kP(X = k)\\
\mu & = \sum_k \dfrac{k(k - 1)(k - 2)}{2} p^3 q^{k - 3}\\
& = \dfrac{p^3}{2}\sum_k k(k - 1)(k - 2) q^{k - 3}\\
\int \mu\, dq & = \dfrac{p^3}{2}\sum_k k(k - 1) q^{k - 2}\\
\iint \mu\, dq\,dq & = \dfrac{p^3}{2}\sum_k k q^{k-1}\\
\iiint \mu\, dq\,dq\,dq & = \dfrac{p^3}{2}\sum_k q^k\\
& = \dfrac{p^3}{2} \left(\dfrac{1}{1 - q}\right)\\
\mu & = \dfrac{p^3}{2}\dfrac{d^3}{dq^3}\left( \dfrac{1}{1 - q} \right)\\
& = \dfrac{p^3}{2}\left( \dfrac{6}{(1 - q)^4} \right)\\
& = \dfrac{p^3}{2}\left( \dfrac{6}{p^4} \right)\\
& = \boxed{\dfrac{3}{p}}
\end{align}$
Thus, the expected number of trials is $\dfrac{3}{p} = \dfrac{3}{0.3} = 10$. Then the expected number of failures is $10 - 3 = 7$.
The events represented by the terms of your summation are not pairwise disjoint. Consider, for example, the outcome in which the first $n$ trials are successes; that situation is counted both in the $k=0$ term and in the $k=1$ term, since the latter includes the sequence of $n$ successes followed by $1$ failure.
We can also use the fact that
$$\sum_{k\ge 0}\binom{n+k}nx^k=\frac1{(1-x)^{n+1}}$$
to observe that if $p=\frac12$, then
$$\sum_{k\ge 0}\binom{n+k}np^n(1-p)^k=\frac1{2^n}\cdot\frac1{(1/2)^{n+1}}=2\;,$$
so your summation must be greater than $1$ for sufficiently large $m$; this doesn’t explain why it’s wrong, but it does show that it can’t be right.
Best Answer
Hint. The (probability that exactly ONE of the failures occurs) is equal to (the probability that the transmission fails and brakes work) plus (the probability that transmission works and brakes fail).