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Question
Six dice thrown 729 times. How many times do you expect at least one die to show 5 or 6?

Answer
Questioner:Akshar
Country:Karnataka, India
Category:Calculus
Private:No
Subject:Probability
Question:Six dice thrown 729 times. How many times do you expect at least one die to show 5 or 6?
A. p(something occurs) = 1.00 – p(it does not)  So:
B. p(at least one die = 5 or 6) = 1 – p(all show 1,2,3,4)
C. p(one die = 1-4) = 2/3.
D. p( multiple independent events ) = product of all the p()’s.
So p(all 6 dice NOT a 5/6) = (2/3)^6 = 64/729
And p(at least 1 5/6) = 1 - (2/3)^ = 1 - 64/729
Now you want an ‘EXPECTATION’ , or ‘EXPECTED VALUE’
Looks like 729 times that probability:  729 * (1 - 64/729)

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