Question

What is the probability of rolling doubles on a pair of dice?

Answer 1

Hint: Here, the pair of dice is thrown simultaneously. We have to evaluate the particular event and then find the favourable outcomes from the assumed sample space and finally evaluate the probability. We are going to follow the same method for solving the given question.
Let us know the concept behind the given question. The rolling doubles means if two coins are flipped, then the outcome should be the same on both coins, it means if the first coin shows head, then the second coin should be identical to the first coin. Then it is said to be doubles.

Complete step by step solution:
If a random experiment is conducted, then each of its outcomes is said to be an event. The set of possible outcomes of a random experiment is called the sample space, and it is denoted as ‘S’.
For an event of pair of dice thrown simultaneously, sample space is given as
\[\begin{align}
 & S=[\left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,4 \right),\left( 1,5 \right),\left( 1,6 \right), \\
 & \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right), \\
 & \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right), \\
 & \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right), \\
 & \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right), \\
 & \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right)] \\
\end{align}\]
Therefore, the number of elements in the sample space is = 36
Hence, for having a double on pair of dice, the favourable outcomes are
\[A=\left[ \left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right),\left( 5,5 \right),\left( 6,6 \right) \right]\]
Therefore, the number of elements in the favourable outcomes = 6
 Then the probability of an event, \[P=\dfrac{Total\text{ }number\text{ }of\text{ }favourable\text{ }outcomes}{total\text{ outcomes}}\]
Hence, the probability of getting a double is
\[P=\dfrac{6}{36}=\dfrac{1}{6}\]

Therefore, the probability of getting a double is \[\dfrac{1}{6}\].

Note: The main concept of solving this question is the basic knowledge of probability. We can also evaluate the sample space by applying the multiplicative identity.