Mixing
Throw a two dices-probability.
$begingroup$
We throw a two symmetric dices. How to show, that we only need to $300$ throws to have $95%$ chance, that in at least $100$ throws we will get a smaller number on the first cube. Is $250$ throws enough?
I thought about Bernoulli trial, where $k=300,n=100$ and probability
of success is equal to $frac{15}{36}$. But these are not calculations for human.
probability
asked Jan 22 at 0:35
pawelKpawelK
547
$endgroup$
add a comment |
$begingroup$
We throw a two symmetric dices. How to show, that we only need to $300$ throws to have $95%$ chance, that in at least $100$ throws we will get a smaller number on the first cube. Is $250$ throws enough?
I thought about Bernoulli trial, where $k=300,n=100$ and probability
of success is equal to $frac{15}{36}$. But these are not calculations for human.
probability
asked Jan 22 at 0:35
pawelKpawelK
547
$endgroup$
3
$begingroup$
the normal approximation to the bernoulli distribution should be good enough. Note: the mean is the number of throws times $frac {15}{36}$. Thus, for $250$ throws the mean is just over $104$ and $sigma approx 7.8$ so there is a very high probability of the first one winning fewer than $100$ times.
$endgroup$
– lulu
Jan 22 at 0:37
add a comment |
$begingroup$
We throw a two symmetric dices. How to show, that we only need to $300$ throws to have $95%$ chance, that in at least $100$ throws we will get a smaller number on the first cube. Is $250$ throws enough?
I thought about Bernoulli trial, where $k=300,n=100$ and probability
of success is equal to $frac{15}{36}$. But these are not calculations for human.
probability
asked Jan 22 at 0:35
pawelKpawelK
547
$endgroup$
We throw a two symmetric dices. How to show, that we only need to $300$ throws to have $95%$ chance, that in at least $100$ throws we will get a smaller number on the first cube. Is $250$ throws enough?
I thought about Bernoulli trial, where $k=300,n=100$ and probability
of success is equal to $frac{15}{36}$. But these are not calculations for human.
probability
probability
asked Jan 22 at 0:35
pawelKpawelK
547
asked Jan 22 at 0:35
pawelKpawelK
547
asked Jan 22 at 0:35
pawelKpawelK
547
asked Jan 22 at 0:35
pawelKpawelK
547
asked Jan 22 at 0:35
pawelKpawelK
547
547
3
$begingroup$
the normal approximation to the bernoulli distribution should be good enough. Note: the mean is the number of throws times $frac {15}{36}$. Thus, for $250$ throws the mean is just over $104$ and $sigma approx 7.8$ so there is a very high probability of the first one winning fewer than $100$ times.
$endgroup$
– lulu
Jan 22 at 0:37
add a comment |
3
$begingroup$
the normal approximation to the bernoulli distribution should be good enough. Note: the mean is the number of throws times $frac {15}{36}$. Thus, for $250$ throws the mean is just over $104$ and $sigma approx 7.8$ so there is a very high probability of the first one winning fewer than $100$ times.
$endgroup$
– lulu
Jan 22 at 0:37
3
3
$begingroup$
the normal approximation to the bernoulli distribution should be good enough. Note: the mean is the number of throws times $frac {15}{36}$. Thus, for $250$ throws the mean is just over $104$ and $sigma approx 7.8$ so there is a very high probability of the first one winning fewer than $100$ times.
$endgroup$
– lulu
Jan 22 at 0:37
$begingroup$
the normal approximation to the bernoulli distribution should be good enough. Note: the mean is the number of throws times $frac {15}{36}$. Thus, for $250$ throws the mean is just over $104$ and $sigma approx 7.8$ so there is a very high probability of the first one winning fewer than $100$ times.
$endgroup$
– lulu
Jan 22 at 0:37
add a comment |
0
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3
$begingroup$
the normal approximation to the bernoulli distribution should be good enough. Note: the mean is the number of throws times $frac {15}{36}$. Thus, for $250$ throws the mean is just over $104$ and $sigma approx 7.8$ so there is a very high probability of the first one winning fewer than $100$ times.
$endgroup$
– lulu
Jan 22 at 0:37