Question

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

\[
  {\text{A}}{\text{. 100}} \\
  {\text{B}}{\text{. 164}} \\
  {\text{C}}{\text{. 170}} \\
  {\text{D}}{\text{. 184}} \\
\]

Answer 1

Hint – To compute the answer, we find the probability of not getting a 6 in a throw with two dice. We use it for n throws and apply log to the equation to get n.
Complete Step-by-Step solution:
The probability of getting a 6 in a single throw of a dice = $\dfrac{1}{6}$

The probability of getting a 6 in a throw with two dice p = $\dfrac{1}{6}$ x $\dfrac{1}{6}$ = $\dfrac{1}{{36}}$

Therefore, the probability of not throwing a double six in one throw with two dice q = 1 – $\dfrac{1}{{36}}$ = $\dfrac{{35}}{{36}}$.

So, the probability of not throwing a double six in any of the n throws = ${{\text{q}}^{\text{n}}}$
Hence the probability of throwing a double six at least once in n throws
 =1 - ${{\text{q}}^{\text{n}}}$ = 1 − ${\left( {\dfrac{{35}}{{36}}} \right)^{\text{n}}}$

Given Data – The probability of at least one double six being thrown in n throws with two ordinary dice is greater than 99%
⟹1 − ${\left( {\dfrac{{35}}{{36}}} \right)^{\text{n}}}$> 0.99
⟹${\left( {\dfrac{{35}}{{36}}} \right)^{\text{n}}}$< 0.01
Apply Log on both sides of the equation,
⟹log ${\left( {\dfrac{{35}}{{36}}} \right)^{\text{n}}}$< log (0.01)
⟹n log $\left( {\dfrac{{35}}{{36}}} \right)$< log (0.01) (log (${{\text{a}}^{\text{x}}}$) = x log a)
⟹n (log 35 – log 36) < log (0.01) (log ($\dfrac{{\text{a}}}{{\text{b}}}$) = log a – log b)
Substitute the log values
⟹n (1.5441−1.5563) < −2
⟹ (−0.0122) n<−2
⟹0.0122n > 2
⟹n > 0.01222​= 163.9
So, the least value of n is 164.
Option B is the correct answer.

Note: The key in solving such types of problems is to find the probability of getting a 6 in a throw with two dice for finding the probability of not throwing a double six in one throw with two dice. Applying this to n throws gives an equation in n. We use logarithms to solve this equation, which is an important step. The log values of decimals can be obtained from the logarithmic table.