A bit (meaning binary digit) is 0 or 1. An ordered array of eight bits (such as01101001) is a byte. How many different bytes are there? If you select a byte at random, what is the probability that you select 11000010? What is the probability thatyou select a byte containing three 1’s and five 0’s?

An ordered array of 8 bits is provided

When the order of arrangement is definite, the permutation is applied and when the order is not definite,combination is applied.

There are 8 digits that are to be filled with 0 or 1, this implies that the total different arrays that can be formed is$2^B$.

There are 8 digits that are to be filled with 0 or 1, this implies that each place has an equal probability of getting filled by 0 or 1 and each digit is independent of other, this implies that probability of selecting 11000010 is$2^{-B}$.

There are three 1s and five 0s and thus the possible arrangements can be $\frac{8!}{3!5!}$.

This implies that the number of outcomes favorable are $\frac{8!}{3!5!}$and total number of outcomes are $2^B$.

Apply the formula for probability, that is $p=\frac{numberofoutcomesfavorabletoE}{totalnumberofoutcomes}$to get the probability of selecting a byte containing three 1’s and five 0’s.

Hence the desired probability is$\frac{7}{32}$.