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What's the correct way to solve following question: Dima has to pick 4 digit password (1-9 inclusive). Every digit is picked randomly. What's the probability of picking at least 1 even digit? Thank you

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  • $\begingroup$ How many possible 4 digits passwords are there? How many of these have only odd digits? $\endgroup$
    – jwc845
    Nov 25, 2019 at 18:50

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You can do this with complementary counting. The probability of picking at least 1 even digit complements the probability of picking all odd digits, which is $\left (\frac{5}{9}\right)^4 = \frac{625}{6561}$, so the answer is $$1 - \frac{625}{6561} = \boxed{\frac{5936}{6561}}$$

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  • $\begingroup$ Thank you very much! $\endgroup$
    – Liorddd
    Nov 25, 2019 at 19:19

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