Say that a number is an *odd-bit* number if
the count of 1-bits in its binary representation is odd.
Define an *even-bit* number analogously.
Thus $541 = 1000011101_2$ is an odd-bit number,
and $523 = 1000001011_2$ is an even-bit number.

Are there, asymptotically, as many odd-bit primes as even-bit primes?

For the first ten primes, we have $$ \lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101 \rbrace $$ with 1-bits $$ \lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace $$ and so ratio of #odd to $n$ is $5/10=0.5$ at the 10-th prime. Here is a plot of this ratio up to $10^5$:

^{ (Vertical axis is mislabeled: It is #odd/$n$.) }

I would expect the #odd/$n$ ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are odd might bias the ratio. The above plot does not suggest convergence by the 100,000-th prime (1,299,709).

Pardon the naïveness of my question.

**Addendum**: Extended the computation to the $10^6$-th prime (15,485,863), where it still
remains 1.5% above $\frac{1}{2}$:

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