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 exactly one 0. a. 000, 001, 010, 011, 100, 101, 110, 111; the probability is $\frac{7}{8}$. b. 001, 011, 101, 111; the probability is $\frac{2}{8}$. c. 000, 010, 011, 101, 111; the probability is $\frac{2}{8}=\frac{1}{4}$. d. 000, 001, 010, 011, 100, 101, 110, 111; the probability is $\frac{3}{8}$.

Explanation:

Step1: Find sample - space size

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

Step2: Count sequences with exactly one 0

The sequences with exactly one 0 are 011, 101, 110. So the number of favorable outcomes $n(A)=3$.

Step3: Calculate probability

The probability formula is $P(A)=\frac{n(A)}{n(S)}$. Substituting $n(A) = 3$ and $n(S)=8$, we get $P(A)=\frac{3}{8}$.

Answer:

D. 000, 001, 010, 011, 100, 101, 110, 111; the probability is $\frac{3}{8}$