Question

In a throw of a pair of dice, what is the probability of getting a doublet?
A.$\dfrac{1}{3}$
B.$\dfrac{1}{6}$
C.$\dfrac{5}{12}$
D.$\dfrac{2}{3}$

Answer 1

Hint: The event of rolling two dice simultaneously is given. We first have to evaluate the sample space of the particular event and then find the favourable outcome from the sample space and finally evaluate the probability. We follow the same methodology for solving the question.

Complete step-by-step answer:

If a random experiment is performed, then each of its outcomes is known as an elementary event. The set of all possible outcomes of a random experiment is called the sample space associated with it and it is generally denoted by ‘S’.
Similarly, for an event of two dice thrown simultaneously, sample space of two dice are given as:
$\begin{align}
  & S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\
 & \text{ }(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
 & \text{ }(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\
 & \text{ }(4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\
 & \text{ }(5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\
 & \text{ }(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \\
\end{align}$
Therefore, total number of elements in sample space = 36
Now, for having a doublet on two dice, the favourable outcomes are given as:
 $F=\left\{ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) \right\}$
Therefore, total number of elements in favourable space = 6
The probability of an event is defined as, $P=\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$.
So, the probability of getting a doublet is:
$P=\dfrac{6}{36}=\dfrac{1}{6}$.
Thus, the probability of getting a doublet is $\dfrac{1}{6}$.

Note: The key concept in solving this problem is the knowledge of probability of a particular event. We can also calculate the sample space by just applying the multiplicative identity. Since, total outcomes of a dice are 6 therefore for two dice the total number of outcomes would be 36. Hence, alternatively without even writing all the cases we can find all the total sample space.