Question

Three identical dice are rolled. The probability that same number will appear on each of them will be
A. \[\dfrac{1}{6}\]
B. \[\dfrac{1}{{36}}\]
C. \[\dfrac{1}{{18}}\]
D. \[\dfrac{2}{{38}}\]

Answer 1

Hint: Probability means possibility of happening of an event. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability of an event can never be negative.

Complete step by step answer:
The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the probability distribution, where you will learn the probability of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
Just as one die has six outcomes and two dice have \[{6^2} = 36\] outcomes. The probability experiment of rolling three dice has \[{6^3} = 216\] outcomes. This idea generalizes further for more dice. If we roll n dice then there are \[{6^n}\] outcomes.
Probability for rolling three dice with the six sided dots such as \[1,2,3,4,5\] and \[6\] dots in each (three) dies.
When three dice are thrown simultaneously/randomly, the number of events can be\[{6^3} = 216\] because each die has \[1\] to \[6\] number on its faces.
.
Three identical dice are rolled. The outcomes that same number will appear on each of them will be as follows
\[\left( {1,1,1} \right),\left( {2,2,2} \right),\left( {3,3,3} \right),\left( {4,4,4} \right),\left( {5,5,5} \right),\left( {6,6,6} \right)\]
All these above cases are considered as favorable outcomes .
We know that Probability (event) \[ = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total number of outcomes}}}\]
Therefore probability(same number appearing on all the three dice) \[ = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total number of outcomes}}}\]
\[ = \dfrac{6}{{216}}\]
Which simplifies to
\[ = \dfrac{1}{{36}}\]

So, the correct answer is “Option B”.

Note: Probability of any event can be between \[0\] and \[1\] only. Probability of any event can never be greater than \[1\] . Probability of any event can never be negative. Here the total number of possible events will be possible outcomes in one attempt with the power of number of attempts.