Question

In a single throw of two dice, the probability of getting a sum of 10 is
A). $\dfrac{1}{12}$
B). $\dfrac{1}{36}$
C). $\dfrac{1}{6}$
D). None of these

Answer 1

Hint: In this question, we will proceed by writing all the possible cases occurring when a dice is thrown . Then we need to find out the cases that will give the sum of 10 in the dice .Then we need to use the probability formula to find the probability.
Formula used :
\[\text{Probability} = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\]

Complete step-by-step solution:
Given, two dice are thrown at the same time.
\[S\ = \{(1,1),(1,2)\ ,(1,3)\ ,(1,4)\ ,(1,5)\ ,(1,6),\ \]
\[\ (2,1)\ ,(2,2)\ ,(2,3)\ ,(2,4)\ ,(2,5)\ ,(2,6),\ \]
\[\ (3,1)\ ,(3,2)\ ,(3,3)\ ,(3,4)\ ,(3,5)\ ,(3,6),\ \]
\[\ (4,1)\ ,(4,2)\ ,(4,3)\ ,(4,4)\ ,(4,5)\ ,(4,6),\ \]
\[\ (5,1)\ ,(5,2)\ ,(5,3)\ ,(5,4)\ ,(5,5)\ ,(5,6),\ \]
\[\ (6,1)\ ,(6,2)\ ,(6,3)\ ,(6,4)\ ,(6,5)\ ,(6,6)\}\]
The probability of total outcomes is \[6^{2}\]
 \[n\left( S \right) = \ 36\]
Let \[A\] be the event of getting a sum of \[10\]
\[A\ = \ \left\{ \left( 6,4 \right),\ \left( 5,5 \right),\ \left( 4,6 \right) \right\}\]
Now,
\[n\left( A \right) = \ 3\]
The probability,
\[P\left( A \right) = \dfrac{n\left( A \right)}{n\left( S \right)}\]
\[P\left( A \right) = \dfrac{3}{36}\]
By simplifying,
We get,
\[P\left( A \right) = \dfrac{1}{12}\]
Thus the probability of getting same number on both dices is \[\dfrac{1}{12}\]

The probability of getting same number on both dices is \[\dfrac{1}{12}\]
So, Option (A). \[\dfrac{1}{12}\] is correct.

Note: We need to calculate the number of favourable and possible outcomes in each case to calculate the probability of each of the given events. We should also be careful that we don’t count the same event repeatedly by mistake. We get only 3 favourable cases as we never get 7 on a dice which should be kept in mind while solving the dice related problems.