We have 10 people, each watching a movie on one of the 7 days of the week. We want the probability that at least two watch on the same day.

Step 1: Interpret the problem
Each person picks a day independently (assuming uniform random choice, 1/7 per day). This is like the birthday problem: $n=10$ "people" and $d=7$ "days."

Step 2: Find the complement probability
The complement event: all 10 people watch on different days.
If $\(10>7\)$, is it possible for all to choose different days?
No - by the pigeonhole principle, with 10 people and 7 days, at least two must share a day.
So the probability that all choose different days $=0$.

Step 3: Conclude
Thus:

$$
\(P(\text { at least two share a day })=1-P(\text { all different })=1-0=1\) .
$$


Step 4: Check options
That's option D. 1.