practice_problems/problems/unit-01/probability-space/19

[giscus[bot]]

practice_problems/problems/unit-01/probability-space/19

https://mids-w203.github.io/practice_problems/problems/unit-01/probability-space/19.html

[jps245]

Hello. Does anyone have the answer to this one? Good question

My logic was the following:
different combinations of letters = 26!
assume 6 spaces are taken up by hamlet and the other 20 by all other letters:
different placements of word hamlet = 21 (hamlet can start in spaces 1 to 21)
different combinations of all non hamlet letters in the string = 20!

1. probability of the word Hamlet appears, ignoring caps = 20! * 21 / 26! = 21! / 26! = 1 / 2223242526 = .000000127
2. n typists to need for .9 probability -> (1 - .000000127)^n = .1 = 18,175,685

Please let me know if you got something different. Not entirely sure on this one. Thanks