Let's solve the problem step-by-step:

We are asked to find the probability that a randomly selected number from the set $\frac{7}{8} \leq x \leq \frac{4}{3}$ is greater than or equal to 1.

Step 1: Define the range

The range of numbers is $\frac{7}{8} \leq x \leq \frac{4}{3}$. We need to split this range into two parts:

• Numbers less than 1 (but greater than or equal to $\frac{7}{8}$).
• Numbers greater than or equal to 1 (but less than or equal to $\frac{4}{3}$).

Step 2: Length of total interval

The total range of numbers is from $\frac{7}{8}$ to $\frac{4}{3}$.

To find the length of this interval: $\text{Total length} = \frac{4}{3} - \frac{7}{8}$ We need to get a common denominator: $\frac{4}{3} = \frac{32}{24}, \quad \frac{7}{8} = \frac{21}{24}$ $\text{Total length} = \frac{32}{24} - \frac{21}{24} = \frac{11}{24}$

Step 3: Length of interval where $x \geq 1$

Now, we need to find the length of the interval where $x \geq 1$, that is, from 1 to $\frac{4}{3}$: $\text{Length where } x \geq 1 = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}$

Step 4: Probability

The probability is the ratio of the interval where $x \geq 1$ to the total interval length: $\text{Probability} = \frac{\frac{1}{3}}{\frac{11}{24}} = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11}$

Thus, the probability that the randomly selected number will be greater than or equal to 1 is $\boxed{\frac{8}{11}}$, which corresponds to option (D).

Would you like further details or have any questions? Here are some related questions to consider:

1. What is the concept of probability in continuous intervals?
2. How do you find the length of intervals on the real number line?
3. How can common denominators help in subtracting fractions?
4. What is the meaning of probability as a ratio of lengths?
5. How does this problem differ from finding probabilities for discrete sets?

Tip: Always ensure to break down complex ranges into simpler intervals when dealing with probabilities in continuous sets.