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I just wanted to verify if my solution to this question is correct: "Rolling 3 dice. First is a 3 or less. What is the probability that the sum after the second is a 5 or less."

My Solution: The probability the sum after the second is five or less is the same as the probability the sum is 5 or 4 or 3 or 2, which implies $$\begin{aligned} P\{\rm{Sum} \leq 5\} = & P\{\rm{Sum} = 2 \mid \rm{First throw} = 1\} \times P\{\rm{First throw} = 1\} + {}\\ & P\{\rm{Sum} = 3 \mid \rm{First throw} = 1\} \times P\{\rm{First throw} = 1\} + {}\\ & P\{\rm{Sum} = 3 \mid \rm{First throw} = 2\} \times P\{\rm{First throw} = 2\} + \ldots \end{aligned}$$

Clearly for each sum, there are $n-1$ different cases with each term being $1/36$. So, is the probability $10/36$ (edit: Is it $9/36$)?

Edit: The answers below make sense, but I am now wondering where the logic was wrong in my solution?

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    $\begingroup$ You are calculating $P\{\rm{Sum} \leq 5\}$ but problem asks for $P\{\rm{Sum} \leq 5 \mid \rm{First throw} \leq 3\}$. $\endgroup$ – Zoran Loncarevic Sep 26 '15 at 15:59
  • $\begingroup$ @ZoranLoncarevic Thank you for clarifying that. $\endgroup$ – Jojo Sep 26 '15 at 16:17
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As simple tree-like diagram will help to get the enumeration: draw an edge from 1 to 1,2,3,4, i.e. (1,1), (1,2), (1,3), (1,4) and so on for two more outcomes on dice one, there's a total of 4+3+2 = 9 of favorable outcomes out of possible 18. What do you get?

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There are only nine successes, because $4-1$ doesn't count, as $4>3$.
There are eighteen possible outcomes with first roll $\leq3$. The question, I guess, asks for the probability the sum$\leq5$ given that the first is $3$ or less, so it is $$\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(A)}$$

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