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Can you please teach me how to solve this? There are other similar questions that I have, but I don't understand how to solve them. Another question is: How many 10-bit binary sequences contain an odd number of 1s?

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  1. They can either end with $0$'s or $1$'s (2 choices). The remaining 8 bits can be either $1$ or $0$ ($2^8$ choices), so you have $2^9$ different possibilities.

  2. Exactly half of them. The total amount of possibilities is $2^{10}$, so half of this is $\frac{2^{10}}{2} = 2^9$.

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