This question concerns bit strings of length six. These bit strings can be divided up into four types depending on their initial
and terminal bit. Thus the types are: 0XXXX0, 0XXXX1, 1XXXX0, 1XXXX1. How many bit strings of length six must you select before you are sure to have at least 6 that are of the same type?
We have to choose 21 strings to be sure we have chosen at least 6 strings of the same type.
Step-by-step explanation:
Since the string type is determined by the initial and terminal bits as understood from the question, then the value of the bits between the initial and terminal bits is of no concern to us.
Now, to be sure you have atleast 6 of the same type, we select each string five times. By doing this, we have already selected 20 strings because we have 4 strings there. Now if you choose any of the string one more time, we are certain that we must have chosen atleast 6 strings that are the same. This means we have to choose 21 strings to be sure we have chosen at least 6 strings of the same type.
You need to form an algebraic equation. 4.5+h=7.3 therefore h= 7.3 - 4.5 when you take 4.5 to the other side it becomes negative therefore the answer is 2.8.
