Suppose the codeword x = 101101 is transmitted over the binary symmetric channel, with symbol error probability p.
What is the probability that the word received has distance at most 1 from x? Justify your answer.
This was my approach: \begin{align*} P &= P(\mathrm{less \; than \; or \; equal \; to \; 1 \; bit \; error}) \\ &= 1-P(\mathrm{greater \; than \;or \;equal \;to \;2\; bit \;errors}) \\ &= 1- (P(2 \;\mathrm{errors}) + P (3 \;\mathrm{errors}) + P (4 \;\mathrm{errors}) + P (5 \;\mathrm{errors}) + P (6\; \mathrm{errors})) \\&=1- \left({6 \choose2}p^2 (1-p)^4 + {6 \choose3} p^3 (1-p)^3 + {6 \choose4} p^4 (1-p)^2 + {6 \choose5} p^5 (1-p) + {6 \choose6}p^6\right) \end{align*}
Then I got stuck and couldn't figure out an approach to go from here. Is this correct or on the right track? Or am I absolutely wrong and overdoing this?
For each position, there is probability $p$ that there is an error in that position, and probability $(1-p)$ that there is not an error in that position. So for example, the probability that there is an error in the first position and no others is $p \cdot (1-p)^{5}$. How many possible ways can you have a single error? As mentioned in the comments, $(1-p)^6$ gives the probability of no errors. This should lead to a solution. (this is much more efficient than computing 1- P(2 or more errors)).