A normal six-sided die is rolled twice. It is known that the sum of the die is 9. What is the probability that the sum of the dice on the second roll is less than 5?
How I understood the question is that we have 4 ways to get sum of "9: 3 + 6, 4 + 5, 5 + 4, 6 + 3". But if we need to get the sum that's less than 5 after the second roll, we only have one option and that is to roll a 1 on the second roll, because with the remaining three, the sum is either equal to or more than 5.
What I mean is that, the first position after the first roll is occupied by one of the numbers which can give the sum of 9:
3 + 6; 4 + 5; 5 + 4; 6 + 3
and that is either a 3 or a 4, or a 5, or a 6. That is why on the second roll we can only roll a 1 if we want the sum to be less than 5.
Is this the correct way? And how would I need to write the solution?
The sum of the two die is 9 so if the first roll was 6 the second roll was must be 3< 5. The probability of that is 1/6. If the roll on the first die was 5 the roll on the second die was 4< 5. The probability of that is 1/6. Those are the only ways we can get less than 5 on the second roll and still get a total or 9.
The probability of this is 1/6+ 1/6= 1/3.