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

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)

Calculus

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All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.**Education/Credentials**

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