The Road To Success

Calculus Level 4

If something has a 1 in n n chance of success, then after n n tries, we expect to have succeeded once.

Approximately, as n n tends to infinity, what is the limit of the probability that with a 1 in n n chance of success, that after n n tries we have at least 1 success?

100 % \approx 100\% 20 % \approx 20\% 40 % \approx 40\% 60 % \approx 60\% 80 % \approx 80\% Limit does not exist 0 % \approx 0\%

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1 solution

On any given try the probability of failure is 1 1 n 1 - \dfrac{1}{n} , so the probability of n n failures in a row is P n = ( 1 1 n ) n P_{n} = \left(1 - \dfrac{1}{n}\right)^{n} , which tends to e 1 = 1 e e^{-1} = \dfrac{1}{e} as n n \to \infty .

Thus the probability 1 P n 1 - P_{n} of at least one success after n n tries as n n \to \infty is 1 1 e 0.632 1 - \dfrac{1}{e} \approx 0.632 , or 60 % \boxed{\approx 60 \text{\%}} .

Mistakenly took the last step to be 1 e n 1-e^{-n} , get wrong answer 100 % 100\% ... Good solution!

Kelvin Hong - 3 years ago

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