how many

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- Aug 25th 2007, 12:58 AMshirelprobability/expected value exercise
how many

- Aug 25th 2007, 02:16 AMCaptainBlack
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 - Aug 25th 2007, 02:20 AMCaptainBlack
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 - Aug 25th 2007, 03:57 PMSoroban
Hello, shirel!

Another (slightly different) explanation . . .

Quote:

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.