Question

In a biased dice, the probability of getting an even number is twice as an odd number. If two such dice are rolled, what is the probability of getting the sum of 9?
(a) $\dfrac{10}{81}$
(b) $\dfrac{18}{81}$
(c) $\dfrac{1}{12}$
(d) $\dfrac{8}{81}$

Answer 1

We solve this problem first by finding the probabilities of getting an even number and an odd number. We know that the dice has 3 even numbers and 3 odd numbers then if $P,{P}'$ are the probabilities of getting even and odd numbers respectively then
$\Rightarrow 3\times P+3\times {P}'=1$
By using the above equation we find the probability of getting an even number and an odd number then we can find the probability of getting sum 9. By using the condition that if $P\left( E \right), P\left( V \right)$ are probabilities of two independent events then the probability of occurring both the events is given as
$\Rightarrow P=P\left( E \right)\times P\left( V \right)$

Complete step-by-step solution
Let us assume that the probability of getting an even number as ${{P}_{1}}$
Let us assume that the probability of getting the odd number is ${{P}_{2}}$
We are given that the probability of getting an even number is twice an odd number.
By converting the above statement into mathematical equation we get
$\Rightarrow {{P}_{1}}=2{{P}_{2}}$
We know that the dice has 3 even numbers and 3 odd numbers. $P,{P}'$ are the probabilities of getting only “one” even and only “one” odd number respectively. Now, we know that the total probability is 1. So, the sum of above probabilities will be 1. Hence, we will have
$\Rightarrow 3\times P+3\times {P}'=1$
By using the above condition to given problem we get
$\Rightarrow 3\left( {{P}_{1}} \right)+3\left( {{P}_{2}} \right)=1$
By substituting the value of ${{P}_{1}}$ in above equation we get

& \Rightarrow 3\left( 2{{P}_{2}} \right)+3{{P}_{2}}=1 \\\ & \Rightarrow {{P}_{2}}=\dfrac{1}{9} \\\ \end{aligned}$$ By substituting the above result in $${{P}_{1}}$$ we get $$\begin{aligned} & \Rightarrow {{P}_{1}}=2\left( \dfrac{1}{9} \right) \\\ & \Rightarrow {{P}_{1}}=\dfrac{2}{9} \\\ \end{aligned}$$ Now let us assume that the probability of getting the sum as 9 as $$P$$ We know that the dice has numbers from 1 to 6 Let us take each condition where we get sum as 9 when two dices are rolled. (i) Getting 3 on the first dice and 6 on the second dice. Let us assume that the probability in this condition as $$P\left( {{E}_{1}} \right)$$ We know that if $$P\left( E \right),P\left( V \right)$$ are probabilities of two independent events then the probability of occurring both the events is given as $$\Rightarrow P=P\left( E \right)\times P\left( V \right)$$ By using the above condition we get the probability of getting 3 on first dice and 6 on second dice as $$\Rightarrow P\left( {{E}_{1}} \right)={{P}_{2}}\times {{P}_{1}}$$ By substituting the required values we get $$\begin{aligned} & \Rightarrow P\left( {{E}_{1}} \right)=\dfrac{1}{9}\times \dfrac{2}{9} \\\ & \Rightarrow P\left( {{E}_{1}} \right)=\dfrac{2}{81} \\\ \end{aligned}$$ (ii) Getting 6 on first dice and 3 on second dice. Let us assume that the probability in this condition as $$P\left( {{E}_{2}} \right)$$ We know that if $$P\left( E \right), P\left( V \right)$$ are probabilities of two independent events then the probability of occurring both the events is given as $$\Rightarrow P=P\left( E \right)\times P\left( V \right)$$ By using the above condition we get the probability of getting 3 on the first dice and 6 on the second dice as $$\Rightarrow P\left( {{E}_{2}} \right)={{P}_{1}}\times {{P}_{2}}$$ By substituting the required values we get $$\begin{aligned} & \Rightarrow P\left( {{E}_{1}} \right)=\dfrac{2}{9}\times \dfrac{1}{9} \\\ & \Rightarrow P\left( {{E}_{1}} \right)=\dfrac{2}{81} \\\ \end{aligned}$$ (iii) Getting 4 on first dice and 5 on second dice Let us assume that the probability in this condition as $$P\left( {{E}_{3}} \right)$$ We know that if $$P\left( E \right),P\left( V \right)$$ are probabilities of two independent events then the probability of occurring both the events is given as $$\Rightarrow P=P\left( E \right)\times P\left( V \right)$$ By using the above condition we get the probability of getting 3 on first dice and 6 on second dice as $$\Rightarrow P\left( {{E}_{3}} \right)={{P}_{1}}\times {{P}_{2}}$$ By substituting the required values we get $$\begin{aligned} & \Rightarrow P\left( {{E}_{3}} \right)=\dfrac{2}{9}\times \dfrac{1}{9} \\\ & \Rightarrow P\left( {{E}_{3}} \right)=\dfrac{2}{81} \\\ \end{aligned}$$ (iv) Getting 5 on first dice and 4 on second dice. Let us assume that the probability in this condition as $$P\left( {{E}_{4}} \right)$$ We know that if $$P\left( E \right),P\left( V \right)$$ are probabilities of two independent events then the probability of occurring both the events is given as $$\Rightarrow P=P\left( E \right)\times P\left( V \right)$$ By using the above condition we get the probability of getting 3 on first dice and 6 on second dice as $$\Rightarrow P\left( {{E}_{4}} \right)={{P}_{2}}\times {{P}_{1}}$$ By substituting the required values we get $$\begin{aligned} & \Rightarrow P\left( {{E}_{4}} \right)=\dfrac{1}{9}\times \dfrac{2}{9} \\\ & \Rightarrow P\left( {{E}_{4}} \right)=\dfrac{2}{81} \\\ \end{aligned}$$ Now, we know that the probability of getting the sum as 9 is the combination of all four parts that is $$\begin{aligned} & \Rightarrow P=P\left( {{E}_{1}} \right)+P\left( {{E}_{2}} \right)+P\left( {{E}_{3}} \right)+P\left( {{E}_{4}} \right) \\\ & \Rightarrow P=\dfrac{2}{81}+\dfrac{2}{81}+\dfrac{2}{81}+\dfrac{2}{81} \\\ & \Rightarrow P=\dfrac{8}{81} \\\ \end{aligned}$$ **Therefore, the probability of getting sum as 9 of given dice is $$\dfrac{8}{81}$$. So, option (d) is the correct answer.** **Note:** Students may make mistakes in the probability of getting an even number and odd number. We know that the dice has 3 even numbers and 3 odd numbers then if $$P,{P}'$$ are the probabilities of getting only ‘one” even and only “one” odd numbers respectively then $$\Rightarrow 3\times P+3\times {P}'=1$$ The above equation gives the correct probability of given dices. We have other condition that is If $$P,{P}'$$ are the probabilities of getting “all” even and “all” odd numbers respectively then $$\Rightarrow P+{P}'=1$$ Here $$P,{P}'$$ are the probabilities of getting all even numbers and all odd numbers. But before condition $$P,{P}'$$ are probabilities of getting “one” even and “one” odd numbers respectively. Students may confuse these two conditions.