How many bit strings of length eight either start with a 0 bit or end with the two bits 00?

It appears to me that 1/2 of all possible strings start with 0, and 1/4 end with 00. So the issue then is how many bit strings of eight digits can be made.

check my thinking.

If the first bit (left most bit) is a 0, then it can be filled in only each of the bit string can be filled in one way and when bit strings of length eight end with the two bits 00. Each of the remaining five position is represent in the bit string can be filled in 2 ways (i.e., either by 0 or 1).

Hence, there are 1×2×2×2×2×2×1×1 = 25 = 32 bit strings of length eight either start with a 0 bit or end with the two bits 00.

if it begins with 0 we have two to the power of 7 ways of choosing the rest, and if it ends in 00 then we have two to the power of 6 ways of choosing the rest. As the question uses the word OR we add these results to get our answer!!

To find the number of bit strings of length eight that either start with a 0 bit or end with the two bits 00, we need to count each case separately and then add them together.

First, let's consider the number of bit strings that start with a 0 bit. Since the first bit is fixed as 0, we have seven remaining bits to choose from. Each of these remaining bits has two possible values (0 or 1). Therefore, there are 2^7 = 128 bit strings that start with a 0 bit.

Next, let's consider the number of bit strings that end with the two bits 00. Similar to the previous case, the last two bits are fixed as 00. So, we have six remaining bits to choose from. Again, each of these remaining bits has two possible values (0 or 1). Thus, there are 2^6 = 64 bit strings that end with the two bits 00.

Now, if we count both cases separately, we will be counting the bit strings that satisfy both conditions twice (those bit strings that both start with a 0 and end with 00). To avoid counting duplicates, we need to subtract the number of bit strings that satisfy both conditions.

For a bit string to satisfy both conditions (start with 0 and end with 00), it must have a length of at least three. The first bit is fixed as 0, the last two bits are fixed as 00, and we have the remaining bits in between. The remaining bits can be chosen in 2^(8-3) = 2^5 = 32 ways.

Therefore, the number of bit strings that satisfy both conditions is 32.

Now, we can calculate the total number of bit strings that either start with a 0 bit or end with the two bits 00 by adding the number of bit strings that start with 0 (128) and the number of bit strings that end with 00 (64) and subtracting the number of bit strings that satisfy both conditions (32):

Total = Number of bit strings that start with 0 + Number of bit strings that end with 00 - Number of bit strings that satisfy both conditions
Total = 128 + 64 - 32
Total = 160

Therefore, there are 160 bit strings of length eight that either start with a 0 bit or end with the two bits 00.