Two dice are rolled and the sum of the face values is less than six. What is the probability that at least one of the dice came up a three?
This is just taking the initial question that I asked and expanding on it: Two dice are rolled and the sum of the face values is six. What is the probability that at least one of the dice came up a three?
Attempt: So since they are requesting that the sum be less than 6 but still have at least one 3 included I felt that the expression should be: $$P(at\ least\ 3| sum < 6) = P(at\ least\ 3| sum = 5) + P(at \ least\ 3| sum = 4)$$ The reason I only have 4,5 is because if we have two dice the only summations that can have a 3 are 4 and 5. This being the case I proceeded: $$P(sum\ of\ 5)= P(5|(3,2))P(3,2) + P(5|(4,1))P(4,1)$$ $$=(1)(\frac{2}{36}) + (1)(\frac{2}{36})$$ Therefore: $$P(at\ least\ 3| sum = 5) = \frac{P(5|(3,2))P(3,2)}{ P(5|(3,2))P(3,2) + P(5|(4,1))P(4,1)}$$ $$=\frac{1}{2}$$ $$P(sum\ of\ 4)= P(4|(2,2))P(2,2) + P(4|(3,1))P(3,1)$$ $$=(1)(\frac{1}{36}) + (1)(\frac{2}{36})$$ Therefore: $$=\frac{P(4|(3,1))P(3,1)}{P(4|(2,2))P(2,2) + P(4|(3,1))P(3,1)}$$
$$ = (1)(\frac{1}{36} + (1)(\frac{2}{36}$$
$$ = \frac{3}{36}$$
And then I was going to sum up these two according to: $$P(at\ least\ 3| sum < 6) = P(at\ least\ 3| sum = 5) + P(at \ least\ 3| sum = 4)$$
But this then means: $$\frac{1}{2} + \frac{2}{3}$$
Which of course is greater than 1. So where did I go wrong?
Best Answer
Your idea of decomposing into the various possible sums smaller than 6 is correct, but you need to ponderate them with the probability of the smaller sum itself occuring, i.e.:
$$P(\text{at least }3\mid\text{sum}<6)=\sum_i P(\text{at least }3\mid\text{sum}=i)\times P(\text{sum}=i\mid\text{sum}<6)$$
As you correctly pointed out, the only two possible cases are $i=4,5$. There are $10$ possible cases with a sum smaller than $6$, and we get: $$P(\text{sum}=5\mid\text{sum}<6)=\dfrac{4}{10}\qquad\text{and}\qquad P(\text{sum}=4\mid\text{sum}<6)=\dfrac{3}{10}$$
Now counting the cases where at least one $3$ occurs: $$P(\text{at least }3\mid\text{sum}=5)=\dfrac{1}{2}\qquad\text{and}\qquad P(\text{at least }3\mid\text{sum}=4)=\dfrac{2}{3}$$
Finally: $$P(\text{at least }3\mid\text{sum}<6)=\dfrac{1}{2}\times\dfrac{4}{10}+\dfrac{2}{3}\times\dfrac{3}{10}=\dfrac{2}{5}.$$
(Note that, of course, this corresponds to counting the outcomes with at least one $3$ - there are $4$ of them - out of the $10$ possible cases with a sum smaller than $6$).