Question

The probability of at least one double-six being thrown in $n$ throws with two ordinary dice is greater than $99$ percent. Calculate the least numerical value of $n$.
Given, $log36=1.5563$ and $log35=1.5441$

Answer 1

Probability of getting a double is $\frac{1}{36}$
Probability of not getting a double is $\frac{35}{36}$
Therefore probability of getting a least one double six in n throw is
$\frac{1}{36}+\frac{35}{36}.\frac{1}{36}+{\left(\frac{35}{36}\right)}^2.\frac{1}{36}+...+{\left(\frac{35}{36}\right)}^{n-1}.\frac{1}{36}=\frac{1}{36}\frac{1-{\left(\frac{35}{36}\right)}^n}{1-\frac{35}{36}}=1-{\left(\frac{35}{36}\right)}^n$
And this exceeds $99$
$1-{\left(\frac{35}{36}\right)}^n>\frac{99}{100}\Rightarrow {\left(\frac{35}{36}\right)}^n<\frac{1}{100}\Rightarrow n>163.93\Rightarrow n=164$