Question

A game is played with a special fair cube die which has one red side, two blue sides and three green sides. The result is the color of the top side after the dice is being rolled. If the die is rolled repeatedly, the probability that second blue result occurs on or before the tenth roll can be expressed in the form of $\dfrac{{{3^p} - {2^q}}}{{{3^r}}}$ where $p,q,r$ are positive integers then find the value of $p + r - q$ .

Answer 1

We have given a special fair die. Also, the probability of the second blue result occurs on or before the tenth roll is expressed as $\dfrac{{{3^p} - {2^q}}}{{{3^r}}}$ . First, we have to determine the values of $p,q,r$ then after we can calculate the value of $p + r - q$ .
For that, first, we determine the probability of the second blue result occurring on or before the tenth roll and then represent the obtained probability as a given expression. To obtain the probability we use Bernoulli trials.

Complete step by step answer:
Let us consider an experiment is repeated $n$ number of times and the probability of success in on trial is $p$. The probability of getting $r$ success in $n$ trial is given as ${}^n{C_r}{p^r}{\left( {1 - p} \right)^{n - r}}$ .
Thus we can find the values of $p,q,r$ and then can obtain the value of the required expression.
Step by step answer:
Step 1: We have given a special fair die which has one red side, two blue dies and three green dies. First we find the probability of occurrence of blue color.
The probability of occurrence of the blue color is given as $P\left( {Blue} \right) = \dfrac{2}{6}$
$\Rightarrow P\left( {Blue} \right) = \dfrac{1}{3}$
The probability of occurrence of non blue color (Red or green) is given as $P\left( {\operatorname{Re} d\,or\,Green} \right) = \dfrac{4}{6}$
$\Rightarrow P\left( {\operatorname{Re} d\,or\,Green} \right) = \dfrac{2}{3}$
Step 2: Now to determine the probability of the second blue result occurs on or before the tenth roll, first, we determine the probability that the second blue result occurs after the tenth roll and subtract this from $1$ .
To determine the probability that the second blue result occurs after the tenth roll there are two possible cases.
In case 1, let us consider that blue dose does not occur in the first ten rolls. For this case, the probability is given by
${}^{10}{C_{10}}{\left( {\dfrac{2}{3}} \right)^{10}}{\left( {\dfrac{1}{3}} \right)^0}$
$\Rightarrow {\left( {\dfrac{2}{3}} \right)^{10}}$
In the second case, let us consider that blue occurs only one time in the first ten rolls. For this case, the probability is given by
${}^{10}{C_1}{\left( {\dfrac{2}{3}} \right)^9}{\left( {\dfrac{1}{3}} \right)^1}$
$\Rightarrow 10{\left( {\dfrac{2}{3}} \right)^9}\left( {\dfrac{1}{3}} \right)$
Now the probability of the second blue result occurs on or before the first ten rolls is given as $1 - {\left( {\dfrac{2}{3}} \right)^{10}} - 10{\left( {\dfrac{2}{3}} \right)^9}\left( {\dfrac{1}{3}} \right)$
Step 3: Now we represent the above probability in the form of $\dfrac{{{3^p} - {2^q}}}{{{3^r}}}$, so we get
$1 - {\left( {\dfrac{2}{3}} \right)^{10}} - 10{\left( {\dfrac{2}{3}} \right)^9}\left( {\dfrac{1}{3}} \right)$
$\Rightarrow 1 - \dfrac{{{2^{10}}}}{{{3^{10}}}} - 10 \times \dfrac{{{2^9}}}{{{3^{10}}}}$
$\Rightarrow \dfrac{{{3^{10}} - {2^{10}} - 10 \times {2^9}}}{{{3^{10}}}}$
$\Rightarrow \dfrac{{{3^{10}} - {2^{10}}\left( {1 + 5} \right)}}{{{3^{10}}}}$
$\Rightarrow \dfrac{{{3^{10}} - {2^{10}}\left( 6 \right)}}{{{3^{10}}}}$
$\Rightarrow \dfrac{{{3^{10}} - {2^{10}}\left( {2 \times 3} \right)}}{{{3^{10}}}}$
$\Rightarrow \dfrac{{3\left( {{3^9} - {2^{11}}} \right)}}{{{3^{10}}}}$
$\Rightarrow \dfrac{{{3^9} - {2^{11}}}}{{{3^9}}}$
Now the value of $p = r = 9$ and the value of $q = 11$
Step 4: Now the value of the expression $p + r - q$ is given as
$p + r - q$
$\Rightarrow 9 + 9 - 11$
$\Rightarrow 7$

So the value of $p + r - q = 7$

Note: The probability of getting success $r$ times in $n$ trials of an independent experiment is given as ${}^n{C_r}{p^r}{q^{n - r}}$ where $p$ is the probability of success and $q$ is the probability of failure of the event.
To determine the probability of the second blue result occurring on or before the tenth roll, we subtract the probability of the second blue result occurring after the tenth roll from 1.