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If we roll $20$ dice, what is the probability of at least three six?

I solved by using $P(A)=1-P(A^C)$ property. Considering the complement of problem, there is a $5/6$ probability of not rolling a six for any given die. So the result is $1-{\left(\frac{5}{6}\right)}^{20}$ Am I thinking wrong? Any help will be appreciated.

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  • $\begingroup$ $1-{\left(\frac{5}{6}\right)}^{20}$ is the probability of at least one $6$ $\endgroup$
    – Henry
    Commented Dec 14, 2020 at 9:13
  • $\begingroup$ You probably need the binomial distribution here $\endgroup$
    – Henry
    Commented Dec 14, 2020 at 9:14

1 Answer 1

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$$1-\binom{20}{0}\Big(\frac{1}{6}\Big)^0\Big(\frac{5}{6}\Big)^{20}-\binom{20}{1}\Big(\frac{1}{6}\Big)^1\Big(\frac{5}{6}\Big)^{19}-\binom{20}{2}\Big(\frac{1}{6}\Big)^2\Big(\frac{5}{6}\Big)^{18}$$

That is

$$1-\text{Probability of zero six's}-\text{Probability of 1 six}-\text{Probability of 2 six's}$$

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  • $\begingroup$ I see , it makes sense now, thanks $\endgroup$
    – softglance
    Commented Dec 14, 2020 at 10:36

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