To evaluate the product \( B = \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) \times \cdots \times \left(1 - \frac{1}{2024}\right) \times \left(1 - \frac{1}{2025}\right) \), we can simplify each term in the product first:

\[
1 - \frac{1}{n} = \frac{n-1}{n}
\]

Thus, the product \( B \) can be rewritten as:

\[
B = \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{3}{4}\right) \times \cdots \times \left(\frac{2023}{2024}\right) \times \left(\frac{2024}{2025}\right)
\]

Notice that this is a telescoping product. In a telescoping product, most terms cancel out. Let's write out the product explicitly to see the cancellation:

\[
B = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{2023}{2024} \times \frac{2024}{2025}
\]

Observe that the numerator of each fraction cancels with the denominator of the next fraction:

\[
B = \frac{1 \cancel{\times 2 \times 3 \times \cdots \times 2023 \times 2024}}{\cancel{2 \times 3 \times \cdots \times 2023 \times 2024} \times 2025}
\]

After cancellation, we are left with:

\[
B = \frac{1}{2025}
\]

Therefore, the value of the product \( B \) is:

\[
B = \frac{1}{2025}
\]