Probability that XOR of an arbitrary number of random bits is 1

Question

How can we say than the XOR between $n$ (uniform, independent) random bits is $1$ with probability $1/2$? For example if we have 4 random bits, we know that the XOR of them will be $1$ with half probability. The same happens if we have 3, or 4 random bits. But if we have $n$ random bits?

Answer 1

Let $D = \bigoplus_{k = 1}^{n-1} b_i$. Then,

\begin{align}P(\bigoplus_{k = 1}^n b_i = 1) &= P(\bigoplus_{k = 1}^n b_i = 1 | D=1)P(D=1) + P(\bigoplus_{k = 1}^n b_i = 1 | D=0)P(D=0) \\ &= \frac12P(D=1) + \frac12P(D=0) \\ &= \frac12 \end{align}

Answer 2

The last bit is $0$ or $1$, each with probability $\frac12$, independently of the XOR of the previous bits,

so leaves that result the same or changes it, each with probability $\frac12$,

meaning the overall XOR is $0$ or $1$, each with probability $\frac12$.

Answer 3

It's One if number of 1s is Odd. So ignore the first n-1 bits. (their parity is completely random*) Now the last bit determine the final answer (based on number of 1s in the first n-1 bits)
*: Probability that number of 1s in the first n-1 bits is Odd, is 1/2. Try to prove this part yourself.
hint: You can find a 1-to-1 correspondence between even and odd parities. for example for every sequence with even parity there is sequence of bits with odd parity (by switching the last bit) and vice versa.