Question

A die is thrown twice. What is the probability that:
i) 3 will come up either time?
ii) 3 will come up at least once?

Answer 1

Hint: Whenever, we say that a die is thrown twice, it is treated as the same experiment as when two dies are thrown together. Therefore, the sample space for both the experiments will be the same.

Complete Step-by-Step solution:
When we throw a die, there are 6 outcomes. When we throw 2 dies together, there are 6x6=36 outcomes.
But, remember whenever we say that a die is thrown twice, it is treated as the same experiment as when two dies are thrown together. Therefore, the sample space for both the experiments will be the same.
In probability theory, the sample space of an experiment or random trial is the set of all possible outcomes or results of that experiment.
When we throw a single die, the sample space is S = {1, 2, 3, 4, 5, 6}
And, when a die is thrown twice, the sample space is:
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)}.
So, this sample space shows us the possibilities of the outcomes that can come when a die is thrown twice.
For the part (i), we have to find the probability that 3 will come either time.
This means that on both the dices, we will get 3.
So, as we can see in the sample space there is only one possibility of (3,3) which is given.
Therefore, its Probability (P) is:
$\begin{align}
  & P=\dfrac{Fav.Outcomes}{TotalOutcomes} \\
 & \Rightarrow P=\dfrac{1}{36} \\
\end{align}$ As the total number of outcomes is 36 and the favorable outcome is only 1.
Hence, the answer of (i) is $\dfrac{1}{36}$
For the part (ii), we have to find the probability that 3 will come up at least once.
This means that 3 can come upon one of the two thrown die or on both.
So, this condition can be fulfilled by these 3 sample spaces: (3,1), (1,3) and (3,3)
Therefore, its Probability (P) is:
$\begin{align}
  & P=\dfrac{Fav.Outcomes}{TotalOutcomes} \\
 & \Rightarrow P=\dfrac{3}{36}=\dfrac{1}{12} \\
\end{align}$ As the total number of outcomes is 36 and the favorable outcome is only 3.
Hence, the answer of (ii) is $\dfrac{1}{12}$

Note: Instead of looking at the calculations, one should always look upon the logic involved in these types of questions. This will also reduce our time and the large calculations involved.