Question

A die has its six faces marked as 0,1,1,1,6,6 two such dice are thrown together and the total score is recorded
i. How many different scores are possible
ii. What is the probability of getting a total of 7

Answer 1

Hint: In this question it is given that a dice has six faces such that one face has 0 marked on it, three faces has 1 marked on them and the remaining two faces has 6 marked on them. Two dice are thrown together so the sample space will have 36 elements as outcomes of this experiment and many elements will be the same due to the same numbers. To calculate the score we have to add the numbers that come at the top face of both dice. For the first part we have to calculate the different possible scores which will come out to be 0,1,2,6,7,12 whereas for the second part we have to find the probability of getting a total of 7 for which we will proceed after constructing the sample space.

Complete step-by-step answer:
In this question we are given that there are two dice such that the numbers on their faces are as the given sequence 0,1,1,1,6,6 and they are thrown together, the method to evaluate the outcome asked by the question is such that we have to add the numbers coming at the top faces of both the dice after they are thrown.
First we are going to construct the sample space of this experiment, the sample space is the set of all possible outcomes of an experiment through which we have to select our favourable outcomes of a particular event related to the experiment and also we have to select the total outcomes which are generally considered all of them given in the sample space.
The sample space of this experiment is denoted by \[S\] is as follows,
\[S=\left\{ \left( 0,0 \right),\left( 0,1 \right),\left( 1,0 \right),\left( 1,1 \right),\left( 0,6 \right),\left( 6,0 \right),\left( 1,6 \right),\left( 6,1 \right),\left( 6,6 \right) \right\}\]
Now by using permutations and combinations we can say that there are six possible outcomes from both the dice separately, so there are total 36 possible arrangements or outcomes that can occur from rolling two dice together, but here we have only nine outcomes because there are some similar outcomes and in set representation we do not write repeating outcomes.
But to represent the total number of outcomes below is the set representation to explain and not to present in exam in this form or set representation.
Hence the hypothetical set containing all the outcomes is represented as,
\[{{S}_{T{{o}_{{}}}understand}}=\left\{ \begin{align}
  & \left( 0,0 \right)one \\
 & \left( 0,1 \right)three \\
 & \left( 1,0 \right)three \\
 & \left( 1,1 \right)nine \\
 & \left( 0,6 \right)two \\
 & \left( 6,0 \right)two \\
 & \left( 1,6 \right)six \\
 & \left( 6,1 \right)six \\
 & \left( 6,6 \right)four \\
\end{align} \right\}\]
Now we need to understand this set in this way such that the outcomes are represented in ordered pairs and the numbers written to the right of them represent the number of times the corresponding outcomes can occur when we perform the experiment, in other words this hypothetical sample space tells the likeliness of occurrence of the outcomes and the probability will be calculated by using this analogy as represented in this hypothetical sample space.
So now coming back to our question we have two parts to solve,
i. In the first part we have to find all the different possible scores which means different sum of the numbers occurring on the top face of both the dice after they are thrown, so we will consider the original sample space which is,
\[S=\left\{ \left( 0,0 \right),\left( 0,1 \right),\left( 1,0 \right),\left( 1,1 \right),\left( 0,6 \right),\left( 6,0 \right),\left( 1,6 \right),\left( 6,1 \right),\left( 6,6 \right) \right\}\]
Now we have to find the different possible sums occurring when dice are thrown together which can be done by adding both the parts of ordered pairs together which represent the outcomes of each die respectively.
Hence the different possible sums are as follows,
\[\begin{align}
  & 0+0=0 \\
 & 0+1=1 \\
 & 1+0=1 \\
 & 1+1=2 \\
 & 0+6=6 \\
 & 6+0=6 \\
 & 1+6=7 \\
 & 6+1=7 \\
 & 6+6=12 \\
\end{align}\]
As there are some sums which are the same and differ in ordered pair position, so there is no point in writing the similar sums again and again.
So the different possible scores are \[0,1,2,6,7,12\].
Hence there are six different possible scores as outcome when two dice are thrown together and their outputs are added to get the score.
ii. Now in the second part of the question we have to find the probability of getting a total of 7 which means when both the dice are thrown together the numbers occurring at the top face of them both shall be added to get the score and it should be 7.
To find the probability we must find the number of favourable outcomes and also the number of total outcomes which can be done by looking at hypothetical sample space constructed by us above which is as follows,
\[{{S}_{T{{o}_{{}}}understand}}=\left\{ \begin{align}
  & \left( 0,0 \right)one \\
 & \left( 0,1 \right)three \\
 & \left( 1,0 \right)three \\
 & \left( 1,1 \right)nine \\
 & \left( 0,6 \right)two \\
 & \left( 6,0 \right)two \\
 & \left( 1,6 \right)six \\
 & \left( 6,1 \right)six \\
 & \left( 6,6 \right)four \\
\end{align} \right\}\]
Now we can say that to get 7 as the score the sum of two outcomes should also be 7 and the possible pairs which can bring 7 when their outcomes are added are \[\left( 1,6 \right)\] and \[\left( 6,1 \right)\]. The number of occurrences of these pairs is written in the hypothetical sample space which is six times for each pair.
Hence the number of favourable outcomes is \[6+6=12\] and the number of total outcomes is 36 as described earlier by permutation and combination.
So now applying the formula for calculating the probability of the event,
\[P\left( A \right)=\dfrac{Numbe{{r}_{{}}}\ o{{f}_{{}}}\ favourable{{e}_{{}}}\ outcomes}{Numbe{{r}_{{}}}\ o{{f}_{{}}}\ tota{{l}_{{}}}\ outcomes}\]
\[P\left( A \right)=\dfrac{12}{36}\]
\[P\left( A \right)=\dfrac{1}{3}\]
Therefore the probability of occurrence of event A which is getting a sum or score of 7 is, \[P\left( A \right)=\dfrac{1}{3}\].

Note: Remember that when evaluating the probability of the second part you should approach the solution from the basic information given in the question or evaluated by you and do not follow the first part to calculate the probability because there are faces with similar numbers which actually changes the probability of occurrence of a number on the top face. Also the hypothetical sample space is just for you to analyze the hidden features of the solution but in exams there is no need to present it in the solution but you can write it in a rough work column. The construction of sample space is important although it hides some elements. The total number of outcomes when two dice are rolled together is 36 and when there are three dice then the total outcomes become 216 by the use of permutation and combination or principle of multiplication. Another example is when three coins are tossed together the total number of possible outcomes is 8 and remember the order in ordered pairs always matters.