Answer:
Option B
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.
