This is not entirely correct. AES encrypts / decrypts data in blocks (in particular, 128-bit blocks). In addition, in CBC mode, the encryption / decryption of the (i + 1) th block depends on the (i) th block.
So, if a random byte falls into the i-th block (let's say for simplicity that the byte does not intersect two blocks), when you go to decrypt the i-th block, this will give you the wrong decryption (that is, a 128-bit block will be wrong). In addition, since the next block was encrypted using the i-th block, the (i + 1) -th block will also decrypt incorrectly (another 128 bits aka 16 bytes). From there, subsequent blocks will be correct (like all previous blocks).
For more information, I read about encryption modes on Wikipedia.
One more thing: changing a random byte, most likely, will not prevent decryption - it just will not give the original text (of course).
Hope this helps!
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