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 $\log 35=1.5441$

Answer 1

To solve this question we recall the definition of probability. First we calculate the probability of getting a double six in one throw with two dice. Also, we calculate the probability of not getting a double six in one throw. Then, by applying these to $n$ throws we calculate the value of $n$.

Complete step by step answer:
As we know that the probability of an event is written as $P(A)$ where, A is an event.
Also we know that if two events are independent then the joint probability is $P(A)\times P(B)$
We have given that two dice are thrown, here two dice are independent.
So, the probability of getting a double six in one throw with two dice will be $p=\dfrac{1}{6}\times \dfrac{1}{6}=\dfrac{1}{36}$
Now, the probability of not getting a double six in one throw with two dice will be
$\begin{aligned} & q=1-p \\\ & q=1-\dfrac{1}{36} \\\ & q=\dfrac{35}{36} \\\ \end{aligned}$
Now the probability of not throwing a double six in any of the $n$ throws will be ${{q}^{n}}$
So, the probability of throwing a double six at least once in $n$ throws will be $1-{{q}^{n}}=1-{{\left( \dfrac{35}{36} \right)}^{n}}$
We have given that the probability of at least one double-six being thrown in $n$ throws with two ordinary dice is greater than $99$ percent
So,

& 1-{{\left( \dfrac{35}{36} \right)}^{n}}>0.99 \\\ & {{\left( \dfrac{35}{36} \right)}^{n}}<0.01..............(i) \\\ \end{aligned}$$ We have to take logarithms on both sides of the equation to solve the equation, then the eq. (i) becomes $$\log {{\left( \dfrac{35}{36} \right)}^{n}}<\log \left( 0.01 \right)$$ As we know that $$\log {{a}^{n}}=n\log a$$ and $\log \dfrac{m}{n}=\log m-\log n$ So, the above equation will be $$n\left( \log 35-\log 36 \right)<\log \left( 0.01 \right)$$ We have given that $\log 36=1.5563$ and $\log 35=1.5441$ Also, we know that $$\log \left( 0.01 \right)=-2$$ When we substitute these values our equation becomes $\begin{aligned} & n\left( 1.5441-1.5563 \right) < -2 \\\ & n\left( -0.0122 \right) < -2 \\\ & 0.0122n > 2 \\\ & n > 163.9 \\\ \end{aligned}$ **So the value of $n$ will be $164$.** **Note:** The key point to solve this question is that we use logarithm to solve this question. We have given the value of logarithm, if not given students can easily obtain the log value of decimals from the logarithm table. Also, use the logarithm to base $10$ which has values greater than $1$.