Question

A pair of dice is thrown once. Find the probability of getting an even number on the first side?

Answer 1

Hint: To solve this type of examples of probability we must know the event and its sample space. So that finding the probability is like a shot. So here two dice are thrown, so the possible events are 36. And out if that the occurring event should have an even number on its first side that is either 2, 4 and 6. So just count the pairs with 2, 4 and 6 on the first side. And there is your solution!
Formula used:
\[{\text{probability}} = \dfrac{{{\text{required outcomes}}}}{{{\text{Total outcomes}}}}\]

Complete step-by-step answer:
Given that two dice are thrown. So the sample space will be,
\[\left[ \begin{gathered}
  \left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right) \\
  \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right) \\
  \left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right) \\
  \left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right) \\
  \left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right) \\
  \left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right) \\
\end{gathered} \right]\]
So there are 36 possible outcomes.
Now we need to find the event or the outcome that is having an even number on its first side.
Now in the sample space above the number before the comma in every bracket is the number on the first side.
So we will check for 2, 4 and 6 in the brackets as the number before comma. So the outcomes are,
\[\left[ \begin{gathered}
  \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right) \\
  \left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right) \\
  \left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right) \\
\end{gathered} \right]\]
So skipping the first, third and fifth row all others are the outcomes we need.
So now the turn of probability using the formula above that,
\[{\text{probability}} = \dfrac{{{\text{required outcomes}}}}{{{\text{Total outcomes}}}}\]
\[P\left( E \right) = \dfrac{{{\text{18}}}}{{{\text{36}}}}\]
On solving to the simplest form we get,
\[P\left( E \right) = \dfrac{{\text{1}}}{2}\]
This is the probability of getting an even number on the first side.

Note: Note that these problems are totally dependent on the sample space and the concept hidden inside the asked or required probability. Like here they asked the even number on the first side so we cannot take the outcome having even number but on the second other side. This is the only thing to understand. Now one may get confused then why we took \[\left( {2,2} \right),\left( {4,4} \right),\left( {6,6} \right)\] since they have even number on other face but they have even number on the first face also. They satisfy the condition so we considered it.