Question

Two unbiased dice are thrown, find the probability that the sum is 8 or greater appears on the first die.

Answer 1

Hint: Here, we can first calculate the total number of outcomes and assume the two events given above and calculate their probabilities. Then use the formula to evaluate the probability of the events.

Formula Used:
We are using two formulas here that are \[P(\dfrac{A}{B}) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\] ,
\[Probability = \dfrac{{Favourable\,{\rm{ }}outcome}}{{Total\,{\rm{ }}outcomes}}\] .

Complete step-by-step answer:
It is given in the question that two unbiased dice are thrown. So, the total possible outcomes are \[{6^2} = 36\] .
Let’s calculate all the possible total outcomes for getting accurate results without any assumption.
\[\begin{array}{l}{\rm{\{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),}}\\{\rm{(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} }}\end{array}\]
Let us consider that P (A) is the probability of getting a sum of both dice faces 8 or greater than 8.
From all the total possible outcomes mentioned about this is all the possible outcomes for A :\[\]\[\left\{ {\left( {2,6} \right),\left( {3,5} \right),\left( {3,6} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)} \right\}\]P(A) = \[\dfrac{{15}}{{36}} = \dfrac{5}{{12}}\]
Let us consider that P (B) is the probability of getting 4 on the first die.
From all the total possible outcomes mentioned about this is all the possible outcomes for B :\[\left\{ {\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)} \right\}\]
P (B) = \[\dfrac{6}{{36}} = \dfrac{1}{6}\]
Let us consider \[P{\rm{ }}\left( {A{\rm{ }} \cap {\rm{ }}B} \right)\] is the probability of getting 4 on the first die and the sum greater than or equal to 8.
Sample space \[\left( {A{\rm{ }} \cap {\rm{ }}B} \right){\rm{ }} = {\rm{ }}\left\{ {\left( {4,{\rm{ }}4} \right),{\rm{ }}\left( {4,{\rm{ }}5} \right),{\rm{ }}\left( {4,{\rm{ }}6} \right)} \right\}\]
\[P\left( {A{\rm{ }} \cap {\rm{ }}B} \right){\rm{ }} = {\rm{ }}\dfrac{3}{{36}} = \dfrac{1}{{12}}\]

The probability that sum of the numbers is greater than or equal to 8 given that 4 was thrown on first die is: \[P(\dfrac{A}{B}) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\] .
Substituting the values.
\[P(\dfrac{A}{B}) = \dfrac{{\left( {\dfrac{1}{{12}}} \right)}}{{\left( {\dfrac{1}{6}} \right)}}\]
\[ \Rightarrow \dfrac{1}{2}\]

Note: In these types of questions, we should first calculate all the possible outcomes. So, it becomes easy for us to check the probability of all particular occurring events. Carefully check the probability method to use for multiple types of events.