# Intermission

**Computer Science**Level 3

Rows in my theater are made of some number of seats, densely packed from the aisle to the wall, thus having only one exit. As such, when someone needs to get up and out, everyone between that person and the aisle is also forced to get up and out - this can get very frustrating for customers in long rows! So, I've done some analysis, and here's what I've come up with:

- At intermissions, each person in the theater has a 1% chance of needing to get up for something, independent of all other factors. We call these 'the givens'.
- At the start of the intermission, all of the givens get up, and anyone between a given and the aisle must also get up by proxy. We call these customers 'the forced', and one should note that a single customer may be both a given and a forced.
- Customer satisfaction remains good so long as no more than 10% of customers have to get up at these intermissions. This includes both the givens, and the forced. (We include the givens as they may be getting up due to discomfort, or some other reason that is also the theater's fault!)
- Thus, I want to ensure that, for any given intermission, the probability that more than 10% of customers in the theater have to get up is at most 50%.

I can fit 25 rows in my new theater. I want to make the rows as long as possible. How many seats long can I make all these rows, and still meet the customer satisfaction requirements I just outlined? All the rows must be the same size, of course.

Assume the theater is always full, don't worry about empty seats.

You should round down when computing 10% of customers. For a theater with 5 seats per row, 10% of customers = 12, not 13.

###### Image Credit: http://musichalloficial.blogspot.com/2014/08/cine-blog.html

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