Question

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

• A. $100$
• B. $164$
• C. $170$
• D. $184$

Answer 1

Answer: (B)

$164$

The probability 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 $n$ thrown with two dice

$=(q+p{)}^n-q^n$
$=1-q^n=1-{\left(\frac{35}{36}\right)}^n$

$\therefore 1-{\left(\frac{35}{36}\right)}^n>0.99$
$⟹{\left(\frac{35}{36}\right)}^n<0.01$
$⟹n(\log 35-\log 36)<\log 0.01$
$⟹n[15441-15563]<-2$

$⟹-0.0122n<-2$
$⟹0.0122n>2⟹n>163.9$

So, the least value of $n$ is $164$.