Question about Cipher Block Chaining(CBC) I'm studying at the moment the CBC Encryption method,and I was asked in an excercise how many bits would be wrong after decryption if during the transmission 2 bits are interchanged.
My guess is that the whole bits are wrong due to how CBC works, am I correct?
It depends on the position of the two interchanged bits. Assume they have different values, i.e. one is $0$ and the other is $1$. Indeed, if they had the same value, interchanging would not affect plaintext decryption.

Let's say $n$ is the length of every block.





In CBC, every ciphertext block is involved in two plaintext blocks decryption: its own and subsequent one, therefore if one CT is corrupted only two PT will be so.



To move back to your question:

If $b_0$ and $b_1$ fall in the same block


2 plaintext blocks will be corrupted, $n+2$ bits will be corrupted.



If $b_0$ and $b_1$ fall in two adjacent blocks


3 plaintext blocks will be corrupted, $2n+1$ bits affected.



If $b_0$ and $b_1$ fall in two separate blocks


4 plaintext blocks will be corrupted, $2n+2$ bits affected.