Question

Two unbiased dice are thrown 3 times. Find the probability that the sum nine would be obtained once.

Answer 1

Hint: Consider n as the number of times the two dice is thrown, r as the number of times the sum on the two dice turns out to be 9, p as the probability of getting the sum on the two dice equal to 9 in a single throw and q as the probability of not getting the sum on the two dice equal to 9 in a single throw. Now, use the binomial probability formula given as $P\left( r \right){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$ and substitute the value n = 3, r = 1.

Complete step-by-step answer:
Here we have been given the number of time two dice is thrown and we are asked to find the probability of getting the sum on the two dice equal to 9 once. Here we will use the binomial probability to get the answer.
Now, let us consider the number of times the die is thrown is n, r is the number of times the sum on the two dice turns out to be 9, p is the probability of getting the sum on the two dice equal to 9 in a single throw and q as the probability of not getting the sum on the two dice equal to 9 in a single throw. Clearly, this is a binomial probability case and the required probability is given as $P\left( r \right){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$. Let us find the value of p and q.
Now, when two dice is thrown a single time there are 36 possible set of outcomes that we record, they are shown in the table below.

(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(6,6)
(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

From the above table we can see that there are four pairs in which we will get the sum equal to 9, they are: - (3, 6), (4, 5), (5, 4) and (6, 3). Therefore, the number of favourable outcomes is 4 and total number of outcomes is 36. So we get,
$\Rightarrow $ Probability of getting the sum equal to 9 = p = $\dfrac{4}{36}=\dfrac{1}{9}$.
$\Rightarrow $ Probability of not getting the sum equal to 9 = q = $\left( 1-\dfrac{1}{9} \right)=\dfrac{8}{9}$.
Therefore substituting the values n = 3, r = 1, $p=\dfrac{1}{9}$ and $q=\dfrac{8}{9}$ in the expression for the binomial probability we get,
$\begin{align}
  & \Rightarrow P\left( 1 \right){{=}^{3}}{{C}_{1}}{{\left( \dfrac{1}{9} \right)}^{1}}{{\left( \dfrac{8}{9} \right)}^{3-1}} \\
 & \Rightarrow P\left( 1 \right){{=}^{3}}{{C}_{1}}{{\left( \dfrac{1}{9} \right)}^{1}}{{\left( \dfrac{8}{9} \right)}^{2}} \\
\end{align}$
Using the formula $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\begin{align}
  & \Rightarrow P\left( 1 \right)=3\left( \dfrac{1}{9} \right)\left( \dfrac{64}{81} \right) \\
 & \therefore P\left( 1 \right)=\dfrac{64}{243} \\
\end{align}$
Hence, the above relation is our answer.

Note: You must remember the binomial probability formula to solve the above question. Remember the notations and the meanings of the terms like p and q. You may use different notations however it will be easy if you will go with the conventions to avoid confusion. Always remember that if one dice is thrown multiple times then the total outcome increases with the relation ${{6}^{x}}$ but when 2 dice is thrown then it increases with the relation ${{36}^{x}}$.