# Two unbiased dice are thrown. The probability that the total of the numbers on the

Dice is greater than $10$ is $\dfrac{1}{x}$, what is the value of $x$.

Last updated date: 21st Mar 2023

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Hint: Here two unbiased dice are thrown so the total number of possible outcomes will be $36$. You have to count the number of favorable outcomes where the total of the numbers on the dice is greater than $10$.

As we know, when two unbiased dice are thrown total number of outcomes will be $36$, which are

$

\left( {1,1} \right)\left( {1,2} \right)\left( {1,3} \right)\left( {1,4} \right)\left( {1,5} \right)\left( {1,6} \right) \\

\left( {2,1} \right)\left( {2,2} \right)\left( {2,3} \right)\left( {2,4} \right)\left( {2,5} \right)\left( {2,6} \right) \\

\left( {3,1} \right)\left( {3,2} \right)\left( {3,3} \right)\left( {3,4} \right)\left( {3,5} \right)\left( {3,6} \right) \\

\left( {4,1} \right)\left( {4,2} \right)\left( {4,3} \right)\left( {4,4} \right)\left( {4,5} \right)\left( {4,6} \right) \\

\left( {5,1} \right)\left( {5,2} \right)\left( {5,3} \right)\left( {5,4} \right)\left( {5,5} \right)\left( {5,6} \right) \\

\left( {6,1} \right)\left( {6,2} \right)\left( {6,3} \right)\left( {6,4} \right)\left( {6,5} \right)\left( {6,6} \right) \\

$

These are total $36$ outcomes which will be obtained on throwing two unbiased dice.

Let $E = $Event of getting the total of numbers on the dice is greater than $10$.

So total number of outcomes which sum is greater than $10$are $\left( {5,6} \right)\left( {6,5} \right)\left( {6,6} \right)$ and hence favorable outcomes to $E$ $ = 3$

Probability $P\left( E \right) = \dfrac{{\left( {{\text{no}}{\text{. of favorable outcomes}}} \right)}}{{\left( {{\text{total no}}{\text{. of possible outcomes}}} \right)}}$

$\therefore P\left( E \right) = \dfrac{3}{{36}} = \dfrac{1}{{12}}$

On comparing it with $\dfrac{1}{x}$ we get $x = 12$ is the required answer.

Note: Whenever you get this type of question the key concept of solving is you have to know all the outcomes obtained from the event or if you can write them then you must write it on your copy then count all the favorable outcomes and use the formula of probability to get the answer.

As we know, when two unbiased dice are thrown total number of outcomes will be $36$, which are

$

\left( {1,1} \right)\left( {1,2} \right)\left( {1,3} \right)\left( {1,4} \right)\left( {1,5} \right)\left( {1,6} \right) \\

\left( {2,1} \right)\left( {2,2} \right)\left( {2,3} \right)\left( {2,4} \right)\left( {2,5} \right)\left( {2,6} \right) \\

\left( {3,1} \right)\left( {3,2} \right)\left( {3,3} \right)\left( {3,4} \right)\left( {3,5} \right)\left( {3,6} \right) \\

\left( {4,1} \right)\left( {4,2} \right)\left( {4,3} \right)\left( {4,4} \right)\left( {4,5} \right)\left( {4,6} \right) \\

\left( {5,1} \right)\left( {5,2} \right)\left( {5,3} \right)\left( {5,4} \right)\left( {5,5} \right)\left( {5,6} \right) \\

\left( {6,1} \right)\left( {6,2} \right)\left( {6,3} \right)\left( {6,4} \right)\left( {6,5} \right)\left( {6,6} \right) \\

$

These are total $36$ outcomes which will be obtained on throwing two unbiased dice.

Let $E = $Event of getting the total of numbers on the dice is greater than $10$.

So total number of outcomes which sum is greater than $10$are $\left( {5,6} \right)\left( {6,5} \right)\left( {6,6} \right)$ and hence favorable outcomes to $E$ $ = 3$

Probability $P\left( E \right) = \dfrac{{\left( {{\text{no}}{\text{. of favorable outcomes}}} \right)}}{{\left( {{\text{total no}}{\text{. of possible outcomes}}} \right)}}$

$\therefore P\left( E \right) = \dfrac{3}{{36}} = \dfrac{1}{{12}}$

On comparing it with $\dfrac{1}{x}$ we get $x = 12$ is the required answer.

Note: Whenever you get this type of question the key concept of solving is you have to know all the outcomes obtained from the event or if you can write them then you must write it on your copy then count all the favorable outcomes and use the formula of probability to get the answer.

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