You roll two standard number cubes. what is the probability that the sum is odd, given than one of the number cubes shows a 1?

 It's not specified whether 1 is the 1st or 2nd roll: HOWER:

The 1st Roll is "1": P(odd sum/the 1st Roll is 1)
What is the sample space of all numbers starting with "1":
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),} = 6
the couple of add sum=(1,2), (1,4), (1,6), =3
P(odd sum/ 1st is 1) = 3/6 =1/2
or in applying the formula:

P(odd sum/the 1st Roll is 1) =P(odd sum ∩ 1) / P(getting "1") it will give the same probability = 1/2

NOW if the 2nd Roll is "1", it 's still 1/2