Final answer:
There are 47 bit strings of length 8 that contain three consecutive 0s or four consecutive 1s.
Step-by-step explanation:
To find the number of bit strings of length 8 that contain three consecutive 0s or four consecutive 1s, we can use the principle of inclusion-exclusion. Let's first calculate the number of bit strings that have three consecutive 0s.
If we fix three consecutive 0s in a string, we are left with 5 remaining positions. Each of these positions can be either a 0 or a 1, giving us 2 choices for each position. So, there are 2^5 = 32 bit strings that have three consecutive 0s.
Now let's find the number of bit strings that have four consecutive 1s. Similar to the previous case, if we fix four consecutive 1s in a string, we are left with 4 remaining positions. Each of these positions can be either a 0 or a 1, giving us 2 choices for each position. So, there are 2^4 = 16 bit strings that have four consecutive 1s.
However, we have counted some bit strings twice, those that have both three consecutive 0s and four consecutive 1s. To find the total number of bit strings that satisfy the given condition, we can subtract the number of overlapping bit strings from the total. In this case, there is exactly one bit string that has both three consecutive 0s and four consecutive 1s. Therefore, the total number of bit strings that contain three consecutive 0s or four consecutive 1s is 32 + 16 - 1 = 47.