each die has 6 faces : total outcomes=6*6=36 there is 5 outcomes where sum=6

dice must be different

remove where both dice are same (3,3) is not allowed.

outcomes=4 (1,5),(2,4),(4,2),(5,1)

p=outcomes/total outcomes =4/36 = 1/9

total pairs are 36

only 4 pairs, (1,5),(5,1),(2,4),(4,2) satisfy two being not same and total is 6

so,4/36

prob=1/9

Great Question

can anyone tell me where we need to write the code i am only getting an option to tick the correct answer from MCQ.

There are 6 possibilities on each die. On 2 dice, there are 6 * 6 = 36 possibilities

There are 4 cases that match the desired criteria: (1,5) (5,1) (2,4) (4,2)

This gives us a probability of 4/36 = 1/9