Question 1194661

When two dice are rolled together then total outcomes are {{{36}}} and 
Sample space is 

[ (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, pairs with sum {{{9}}} are ({{{3}}},{{{ 6}}}) ({{{4}}}, {{{5}}}) ({{{5}}}, {{{4}}}) ({{{6}}}, {{{3}}}) i.e. total {{{4}}} pairs

Total outcomes = {{{36}}}
Favorable outcomes = {{{4}}}

Probability of getting the sum of 9 = Favorable outcomes / Total outcomes
                                                       = {{{4 / 36 = 1/9}}}

So, P(sum of {{{9}}}) ={{{ 1/9}}}.


answer:
c.
{{{1/9}}}