Question

The probability that in the toss of two dice we obtain the sum 7 or 11 is
A) $ \dfrac{1}{6} $
B) $ \dfrac{1}{{18}} $
C) $ \dfrac{2}{9} $
D) $ \dfrac{{23}}{{108}} $

Answer 1

Hint: We will first find the total favorable outcome that has the sum as 7 after that we will find the same for 11 by finding the numbers that are added up to give 11 as outcome. Then we will add both the outcomes that will be the total number of favorable outcomes and we will use the probability formula that will be the final answer respectively.
Formula used:
Is S is the set of all the outcomes and A is the set of favorable outcomes, then the probability of occurrence of A is given by,
 $ P(A) = \dfrac{{n(A)}}{{n(S)}} $

Complete step-by-step answer:
The possibility of outcomes that we will get while throwing two dices is,
 $ S = \left\{ {(x,y)|x,y = 1,2,3,4,5,6} \right\} $
Let’s list the elements they are $ S = \left\{ \begin{gathered}
  (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) \\
\end{gathered} \right\} $
The required outcome according to the question is the sum of outcomes of the dices are 7 or 11
Let set A contains all such outcomes where the sum of outcomes of two die is 7,
So, $ A = \left\{ {(x,y)|x + y = 7} \right\} $
Let’s list the elements,
 $ \Rightarrow A = \left\{ {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} \right\} $
Probability of getting sum of outcomes of two die as 7 is,
 $ \Rightarrow P(7) = \dfrac{{n(A)}}{{n(S)}} = \dfrac{6}{{36}} $ --(1)
Let set B contains all such outcomes where the sum of outcomes of two die is 7,
So, $ B = \left\{ {(x,y)|x + y = 11} \right\} $
Let’s list the elements,
 $ \Rightarrow B = \left\{ {(5,6),(6,5)} \right\} $
Probability of getting sum of outcomes of two die as 11 is,
 $ \Rightarrow P(11) = \dfrac{{n(B)}}{{n(S)}} = \dfrac{2}{{36}} $ --(2)
Adding (1) and (2) we get
 $ \Rightarrow P(7) + P(11) = \dfrac{6}{{36}} + \dfrac{2}{{36}} $
 $ \Rightarrow P(7) + P(11) = \dfrac{{6 + 2}}{{36}} = \dfrac{8}{{36}} $
Thus probability of getting sum of outcomes of two die as 11 or 7 is
 $ \Rightarrow P(A) + P(B) = \dfrac{8}{{36}} = \dfrac{2}{9} $
So, the correct answer is “Option C”.

Note: Always remember that in total we have 36 outcomes for 2 dice rolled at a time and for a single dice we have only 6 outcomes. And if probability of occurrence of two events is known then probability of occurrence of any one of them is the sum of probabilities of both of them.