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 $\log 36=1.5563$ and $\log 35=\mathrm{1.5441.}$

• A. The least value of n is $163$
• B. The least value of n is $164$
• C. The least value of n is $162$
• D. The least value of n is $165$

Answer 1

Answer: (B)

The least value of n is $164$

Probability of getting a double six in one throw with two dice $p=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$
$\therefore$ The probability of not throwing a double six in one throw with two dice $q=1-\frac{1}{36}=\frac{35}{36}$
So, the probability of not throwing a double six in any of the n throws=$q^n$
Hence the probability of throwing a double six at least once in n throws =$1-q^n=1-{\left(\frac{35}{36}\right)}^n$
Now according to the question,
$1-{\left(\frac{35}{36}\right)}^n>0.99$
$\Rightarrow {\left(\frac{35}{36}\right)}^n<0.01$ ...(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)
or $n[lo{g}_{10}35-lo{g}_{10}36]<lo{g}_{10}0.01$
or $n[1.5441-1.5563]<-2$
or $-0.0122n<-2$
or $0.0122n>2$
or $n>\frac{2}{0.0122}=163.9$
So, the least value of n is $164.$