Given that there are 32 eight-bit strings that begin 101 and 16 eight-bit strings that begin 1101, how many eight-bit strings begin either 101 or 1101?

naivlingr

naivlingr

Answered question

2021-02-25

Given that there are 32 eight-bit strings that begin 101 and 16 eight-bit strings that begin 1101, how many eight-bit strings begin either 101 or 1101?

Answer & Explanation

gotovub

gotovub

Skilled2021-02-26Added 98 answers

Let A represent the collection of all eight-bit strings that start with 101, and B represent the collection of all eight-bit strings that start with 1101.
n(A)=32,

n(B)=16

(n(X) tells us how many elements there are in X.)
A and B are not connected, hence we obtain that

n(AB)=n(A)+n(B)=32+16=48,

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?