With all respect, never ask a mathematician to solve an engineering problem; the lack of precision inherent in the real world gives them heart palpitations:-)
As the question states, " to a reasonable approximation ... 10% chance of losing the whole packet" (mathematicians can't read either:-)). There is also an inherent assumption that the loss of fragments is independent; in a real network this is unlikely to be the case.
So the chance of losing both is $.1\times.1=.01$. So the chance of getting at least 1 is $1-0.01=0.99$.
For part B, the chance of losing any single fragment is $0.01\times0.01=0.0001$. Now, if any of the 10 fragments is lost twice the whole packet is lost. So we must get 10 good fragments, this happens with a probability of $(1-0.0001)^{10}\approx0.999$ which gives a loss probability of $0.001$.
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 $10 - 3 = 7$.
Best Answer
Probability of success in $0.7$. The distribution of number of trials until success is geometric with parameter of $0.7$. So the expected value of trials until success is $\frac{1}{0.7}$.