If this is meant to be practical question, where a good approximation is sufficient, then yes your computation is good! However, it is not the exact answer.
As leonbloy pointed out, the case where all the bits are always transferred correctly isn't the only way for the message to be the same at the end. Let's think about a simpler case, with only one bit and four transfers. Let's say the bit starts as a $1$, and let's write "F" when it flips during a transfer, and "N" when nothing happens. Here are a few examples of possible scenarios :
\begin{align*}
1 \underset{N}{\longrightarrow} 1 \underset{N}{\longrightarrow} 1 \underset{N}{\longrightarrow} 1 \underset{N}{\longrightarrow} 1
\\ \\
1 \underset{N}{\longrightarrow} 1 \underset{F}{\longrightarrow} 0 \underset{N}{\longrightarrow} 0 \underset{N}{\longrightarrow} 0
\\ \\
1 \underset{F}{\longrightarrow} 0 \underset{N}{\longrightarrow} 0 \underset{N}{\longrightarrow} 0 \underset{F}{\longrightarrow} 1
\\ \\
1 \underset{F}{\longrightarrow} 0 \underset{N}{\longrightarrow} 0 \underset{F}{\longrightarrow} 1 \underset{F}{\longrightarrow} 0
\\ \\
1 \underset{F}{\longrightarrow} 0 \underset{F}{\longrightarrow} 1 \underset{F}{\longrightarrow} 0 \underset{F}{\longrightarrow} 1
\end{align*}
As you can see, there's actually many ways for the bit to be the same at the beginning and at the end. It's hard to imagine because it seems unlikely that a large message would coincidentally have the right flips to still be preserved, but it is still a non-negligible part of it.
Thinking about it, you might notice that the bit stays preserved whenever there's an even number of flips happening, so that a flip is always cancelled out by another flip. In our simple case, the probability that this happens is :
$$
\mathbb{P}(\text{"preserved"}) = \mathbb{P}(\text{"0 flips" or "2 flips" or "4 flips"}),
$$
and since those events are all disjoint :
$$
\mathbb{P}(\text{"preserved"}) = \mathbb{P}(\text{"0 flips"}) + \mathbb{P}(\text{"2 flips"}) + \mathbb{P}(\text{"4 flips"}).
$$
The number of flips follows a binomial distribution, and so :
$$
\mathbb{P}(\text{"preserved"}) = \binom{4}{0}(1-p)^4 + \binom{4}{2}p^2(1-p)^2 + \binom{4}{4}p^4 \approx 0.996,
$$
which is $0.006$ higher than just the probability of no flips happening.
Let's generalize even more by considering $T\in\mathbb{N}$ transfers. Now, the probability that a random number following a binomial distribution is even is actually given by the simple formula :
$$
\mathbb{P}(\text{"preserved"})
= \mathbb{P}(\text{"even number of flips"})
= \frac{1}{2}(1+(1-2p)^T).
$$
Several proofs of this expression are given here, as well as this answer I really like.
Let's further consider a message made up of $n$ bits. Since the bits are independent of each other, the probability that the whole message gets preserved is :
$$
\mathbb{P}(\text{"n bits are preserved"})
=\mathbb{P}(\text{"1 bit is preserved"})^n
= \left(\frac{1+(1-2p)^T}{2}\right)^n.
$$
Plugging in our values, $T=10$, $p=0.001$, and $n=100$, we get :
$$
\mathbb{P}(\text{"n bits are preserved"}) \approx 0.369357
$$
which is just about a $0.16$% difference from the no-flips case. Not a whole lot in practice, but notice that over a very large number of transfers $T$, the probability of the message being preserved would approach $1/2^n$, whereas the probability of no flips at all would approach $0$. Quite different!
im not sure how to take into account that if one bit changes in one transfer it could get "fixed" in any of the other answers.
– Noolenne Jul 05 '23 at 18:06but i guess having an answer for both would be nice. The question is from a statistical compendium i've been using to learn probabilities
– Noolenne Jul 05 '23 at 18:56