What is the probability of getting a number on the first throw greater

exactly my approach...yet I used approximations.
the odds should be slightly less than 1/2
A is too little
B is too big
C is close
D is too big
E is too big.

C remains

if in first through prob of getting 1 is 1/6 but in this case other number greater than 1 will be 5/6
similarly prob of getting 2 is 1/6 but in this case other number greater than 2 will be 4/6

adding till 5
1/6( 5/6+ 4/6....1/6)= 5/12

C is the answer

Probability of getting it is 1/2 minus the probability of getting the same number on both throws which is 1/12.

(1/2) - (1/12) = 5/12

Option C

total number of outcomes, 6 +6 (1st and 2nd throw )
outcomes on 1st throw==================================123456
in order for outcome to be greater than 1st throw 2nd throw outcome =====23456

so 5 outcomes of higher value ,
answer 5/12.
option c

Easy to understand method:

Numbers in brackets are numbers rolled in 2nd throw.

First throw equals 1:
Favorable case when 1st throw > 2nd throw is (not possible) = 0.
Unfavorable case when 1st throw <= 2nd throw is (1,2,3,4,5,6). = 6

First throw equal 2:
Favorable case when 1st throw > 2nd throw is (1) = 1
Unfavorable case when 1st throw <= 2nd throw is (2,3,4,5,6) = 5

First throw equal 3:
Favorable case when 1st throw > 2nd throw is (1,2) = 2
Unfavorable case when 1st throw <= 2nd throw is (3,4,5,6) = 4

First throw equal 4:
Favorable case when 1st throw > 2nd throw is (1,2,3) = 3
Unfavorable case when 1st throw <= 2nd throw is (4,5,6) = 3

First throw equal 5:
Favorable case when 1st throw > 2nd throw is (1,2,3,4). = 4
Unfavorable case when 1st throw <= 2nd throw is (5,6) = 2

First throw equal 6:
Favorable case when 1st throw > 2nd throw is (1,2,3,4,5) = 5
Unfavorable case when 1st throw <= 2nd throw is (6) = 1

Probabality = Total Favorable/Total favorable + unfavorable
=15/36
=5/12

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