Question

a computer is programmed to generate a sequence of three digits, where each digit is either 0 or 1, and each of these is equally likely to occur. construct a sample space that shows all possible three - digit sequences of 0s and 1s and then find the probability that a sequence will contain at least one 0.
a. 000, 001, 010, 011, 100, 101, 110, 111; $p(\text{at least one }0)=\frac{7}{8}$
b. 000, 001, 010, 100, 101, 110 ; $p(\text{at least one }0)=\frac{6}{6}$
c. 100, 101, 110, 111; $p(\text{at least one }0)=\frac{3}{4}$
d. 000, 001, 010, 011; $p(\text{at least one }0)=\frac{4}{8}=\frac{1}{2}$

Explanation:

Step1: List all possible sequences

Each digit has 2 choices (0 or 1). For a three - digit sequence, by the multiplication principle, the total number of possible sequences is $2\times2\times2 = 8$. The sample space $S=\{000,001,010,011,100,101,110,111\}$.

Step2: Find the number of sequences with at least one 0

The number of sequences with no 0s is 1 (i.e., 111). So the number of sequences with at least one 0 is $n = 8 - 1=7$.

Step3: Calculate the probability

The probability $P$ of an event is given by $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Here, the total number of outcomes is 8 and the number of favorable outcomes (sequences with at least one 0) is 7. So $P(\text{at least one }0)=\frac{7}{8}$.

Answer:

A. 000, 001, 010, 011, 100, 101, 110, 111; $P(\text{at least one }0)=\frac{7}{8}$