A die is rolled three times. The probability of getting their sum equal to a prime number of the form (4 n+1 ) is

Step-by-step solution

We obtain following prime sum in three throws of dice; 3, 5, 7, 11, 13 and 17 And sum in from of (4 m+1 ) are, 5, 13 and 17 Following are favourable outcomes for above sums ( aligned & (1,2,2),(2,1,2),(2,2,1) & (3,1,1),(1,3,1),(1,1,3) & (5,3,3),(3,5,3),(3,3,5) & (6,6,1),(1,6,6),(6,1,6) & (4,4,5),(5,4,4),(4,5,4) & (2,6,5),(6,2,5),(5,2,6) aligned ) .etc. Total number of out comes of obtaining sum a prime of from ' (4 m+ l ^ = sum 5(6 ) times ()+ sum 13 ) (21 times) + sum 17 ( 3 times) (=30 ) cares. Total number of po