how many
In $\displaystyle N$ trials the number of times, $\displaystyle n$ that face $\displaystyle 5$ shows on a fair dice has a binomial distribution $\displaystyle B(N, 1/6)$.
So your question is asking for the smallest $\displaystyle N$ such that:
$\displaystyle
p(n=0) \le 0.1
$
so:
$\displaystyle
p(n=0) = b(0;N,1/6)=\left(\frac{1}{6}\right)^N \le 0.1
$
so now find N.
RonL
Whether you suspect that a coin is unfair if it fails a test that a fair coin
passes 90% of the time depends on you, do you have some other reasom to
suspect the coin? Otherwise you may suspect it and possibly do somemore tests.
If it fails a 99% test you may have more reason to suspect the coin.
RonL
Hello, shirel!
Another (slightly different) explanation . . .
How many times does one have to roll a die in order
to have face 5 showing up with a 90% certainty?
We know that for one roll: .$\displaystyle P(5) = \frac{1}{6},\;\;P(\text{not 5}) = \frac{5}{6}$
Consider $\displaystyle n$ consecutive rolls without a 5.
. . This probability is: .$\displaystyle \left(\frac{5}{6}\right)^n$
And we want this probability to be less than 10%.
. . Hence, we have: .$\displaystyle \left(\frac{5}{6}\right)^n \:<\:0.1$
Take logs: .$\displaystyle \log\!\left(\frac{5}{6}\right)^n \:<\:\log(0.1) \quad\Rightarrow\quad n\cdot\log\!\left(\frac{5}{6}\right) \;<\;\log(0.1)$
. . Divide by $\displaystyle \log\!\left(\frac{5}{6}\right)$ which is negative: .$\displaystyle n \;> \;\frac{\log(0.1)}{\log\!(\frac{5}{6})} \;\approx\;12.629$
Therefore, one must roll a die at least 13 times.