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

Answer:

Explanation:

 The probabilty of getting a double six in one throw of two dice

$=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

 $\therefore $     $p=\frac{1}{36}$

 q=1- p

 $=1-\frac{1}{36}=\frac{35}{36}$

 Now, (p+q)^m

 $=q^{n}+{^{n}C_{1}}q^{n-1}p+{^{n}C_{2}}q^{n-2}p^{2}+.....+{^{n}C_{r}}q^{n-r}p^{r}+.....+p^{n}$

 the probability  of getting atleast one double six in n throws with two dice.

 $(q+p)^{n}-q^{n}$

 $ 1-q^{n}=1-(\frac{35}{36})^{n}$

$\therefore$   $1-(\frac{35}{36})^{n}>0.99$

$\Rightarrow(\frac{35}{36})^{n}< 0.01$

$\Rightarrow n(\log35-\log36)<\log0.01$

$\Rightarrow n(15441-15563)<-2$

$\Rightarrow -0.0122n<-2$

 $\Rightarrow -0.0122n>-2\Rightarrow n>163.9$

 So, the least value of n is 164.