Question

The probability of at least one double-six being thrown in ‘n’ throws with two ordinary dice is greater than 99percent. Calculate the least numerical value of ‘n’. Given $\log 36=1.5563$ and $\log 35=1.5441$.
(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

Hint: To solve the question given above, we will first find out the probability of a double-six when the dice are thrown a single time and the probability of any other combination when the dice is thrown single time. Then, we will form the cases according to which throw the double-six will come. Then, we will add all the cases and then we will apply the condition given in the question that probability is greater than 99 percent and find the value of ‘n’.

Complete step-by-step answer:
When one dice is thrown then the probability that six will come $=\dfrac{1}{6}$. When two dices are thrown then the probability of double-six$={{\left( \dfrac{1}{6} \right)}^{2}}=\dfrac{1}{36}$.
The probability that any other combination will come $=1-\dfrac{1}{36}=\dfrac{35}{36}$ .
We will denote the probability of double-six by P(T) and the other combination is P(F). Thus, we have:
$P\left( T \right)=\dfrac{1}{36}$
$P\left( F \right)=\dfrac{35}{36}$
Now, we will form cases according to which throw the double-six will come.
Case 1: The double-six comes at the very first throw. The probability of happening this is denoted by $P\left( {{x}_{1}} \right)$. Thus, we have:
$P\left( {{x}_{1}} \right)=P\left( T \right)$
Case 2: The double-six comes at the second throw. This means that at first throw, any other combination will come. The probability of happening this is denoted by $P\left( {{x}_{2}} \right)$. Thus, we have:
$P\left( {{x}_{2}} \right)=P\left( F \right).P\left( T \right)$
Case 3: The double-six comes at the third throw. This means that at first two throws, any other combination will come. The probability of happening this is denoted by $P\left( {{x}_{2}} \right)$. Thus, we have:
$P\left( {{x}_{3}} \right)=P\left( F \right).P\left( F \right).P\left( T \right)$
$\Rightarrow P\left( {{x}_{3}} \right)={{\left[ P\left( F \right) \right]}^{2}}.P\left( T \right)$
Similarly, we can say that if the double-six comes at ${{r}^{th}}$ throw then the probability of happening this is given by:
$P\left( {{x}_{r}} \right)={{\left[ P\left( F \right) \right]}^{r-1}}.P\left( T \right)$
There will be total (x) cases. The last case will be when the double-six comes at the last throw.
Now, the total probability will be given by adding the probabilities from all the cases. Thus, we will get:
Total probability $=P\left( {{x}_{1}} \right)+P\left( {{x}_{2}} \right)+P\left( {{x}_{3}} \right)........+P\left( {{x}_{n}} \right)$
Total probability$=P\left( T \right)+P\left( F \right).P\left( T \right)+{{\left[ P\left( F \right) \right]}^{2}}.P\left( T \right)+...{{\left[ P\left( F \right) \right]}^{n-1}}.P\left( T \right)$
Total probability$=P\left( T \right)\left[ 1+P\left( F \right)+{{\left[ P\left( F \right) \right]}^{2}}+......P{{\left( F \right)}^{n-1}} \right]$
Now, we can see that the above series is a geometric progression. Now the sum of the terms of G.P with first term ‘a’ and the ratio ‘r’ is given by:
${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$
Thus, we have:
Total probability$=P\left( T \right)\left[ \dfrac{\left( 1 \right)\left( 1-{{\left[ P\left( F \right) \right]}^{n}} \right)}{1-P\left( F \right)} \right]$
Now, we will put the values of $P\left( T \right)$ and $P\left( F \right)$ in the above equation. Thus, we will get:
Total probability$=\left( \dfrac{1}{36} \right)\left[ \dfrac{1-{{\left( \dfrac{35}{36} \right)}^{n}}}{1-\left( \dfrac{35}{36} \right)} \right]$
Total probability$=\left( \dfrac{1}{36} \right)\left[ \dfrac{1-{{\left( \dfrac{35}{36} \right)}^{n}}}{\dfrac{1}{36}} \right]$
Total probability$=1-{{\left( \dfrac{35}{36} \right)}^{n}}$
Now, it is given that the probability is greater than 99%. So, we will get:
$1-{{\left( \dfrac{35}{36} \right)}^{n}}>99%$
$\Rightarrow 1-{{\left( \dfrac{35}{36} \right)}^{n}}>\left( \dfrac{99}{100} \right)$
$\Rightarrow 1-\dfrac{99}{100}>{{\left( \dfrac{35}{36} \right)}^{n}}$
$\Rightarrow \dfrac{1}{100}>{{\left( \dfrac{35}{36} \right)}^{n}}$
Now, we will apply the logarithm on both sides. Thus, we will get:
${{\log }_{10}}\left( \dfrac{1}{100} \right)>{{\log }_{10}}{{\left( \dfrac{35}{36} \right)}^{n}}$
Now, we will use the following logarithmic identities:
${{\log }_{a}}{{b}^{n}}>n{{\log }_{a}}b$
$\Rightarrow {{\log }_{10}}{{\left( \dfrac{1}{10} \right)}^{2}}>n{{\log }_{10}}\left( \dfrac{35}{36} \right)$
$\Rightarrow {{\log }_{10}}{{10}^{-2}}>n{{\log }_{10}}\left( \dfrac{35}{36} \right)$
$\Rightarrow -2\left( 1 \right)>n{{\log }_{10}}\left( \dfrac{35}{36} \right)$
$\Rightarrow -2>n\log \left( \dfrac{35}{36} \right)$
Now, we will use another logarithmic identity:
${{\log }_{a}}\left( \dfrac{b}{c} \right)={{\log }_{a}}b-{{\log }_{a}}c$
Therefore, we will get:
$-2>n\left( \log 35-\log 36 \right)$
Now the values of $\log 35$and $\log 36$ are given in question. So we get:
$\Rightarrow -2>n\left( 1.5541-1.5563 \right)$
$\Rightarrow -2>n\left( -0.0122 \right)$
$\Rightarrow n>\dfrac{2}{0.0122}$
$\Rightarrow n>163.93$
Therefore the least value of ‘n’ is 164.
Hence option (b) is correct.

Note: The alternate method to find the total probability is shown below:
The probability that no double-six will come in ‘n’ throws will be equal to ${{\left( P\left( F \right) \right)}^{n}}={{\left( \dfrac{35}{36} \right)}^{n}}$
Now, the total probability that at least one double-six will come =1-probability of no double-six. Thus, we have:
Total probability$=1-{{\left( \dfrac{35}{36} \right)}^{n}}$