Question

The probability of at least one double-six being thrown in n throws with two ordinary dice is greater than 99%. Calculate the least numerical value of n.___

Answer 1

The probability of getting a double six in one throw with two dice
$=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$
The probability of not throwing a double six
$=1-\frac{1}{36}=\frac{35}{36}$
Let $p=\frac{1}{36}$ and $q=\frac{35}{36}$
$\therefore$ The probability of not throwing a double six in any of these n throws $=q^n$
Hence the probability of throwing double six at least once in n throws
$=1-q^n=1-{\left(\frac{35}{35}\right)}^n$
$1-\left(\frac{35}{36}\right)>0.99$
${\left(\frac{35}{36}\right)}^n<0.01$ ............ (1)
Since both sides of (1) are + ive, the inequality will not be affected by taking logarithm to the base 10 (which is greater than 1)
$n[lo{g}_{10}35-lo{g}_{10}36]<lo{g}_{10}0.01$
$n[1.5441-1.55563]<-2-0.0122n<-20.0122n>2n>\frac{2}{0.0122}=163.9$

$\therefore$ The least value of n is 164