Throwing two dice: find prob. sum div. by 3 if div. by 2;

[courteous]

»We simultaneously throw two dice. What is the probability that:
a at least one throw is divisible by 3 [SOLVED]
b the sum of two throws is divisible by 3, if the sum is divisible by 2
c the sum is at least 9, if we know, that the sum is greater than 6
d the difference between throws on each dice is 2, if we know, that the sum is greater than 8«

For b, and using \(\displaystyle P(B \mid A) = \frac{P(A \cap B)}{P(A)}\) for conditional probability (of two occurrences: \(\displaystyle B\) sum is divisible by 3, and, \(\displaystyle A\) sum is divisible by 2), where occurrence \(\displaystyle B\) is under the condition of \(\displaystyle A\) occurring. "Sense" says that \(\displaystyle P=\frac{\frac{6}{36}}{\frac{18}{36}}\), that there are half of possible sums divisible by 2, and that there is one third of those that are at the same time divisible by 3 (\(\displaystyle {P(A \cap B)\)).
I need help with rigorously finding the number of each (\(\displaystyle {P(A)\): divisible by 2 and \(\displaystyle {P(A \cap B)\): divisible by 2 and 3).

c: I can list all pairs of throws that give sum greater than 6 (7 or more) ...
sum 7: 1,6 - 2,5 - 3,4 (all multiplied by 2)
sum 8: 2,6 - 3,5 (all multiplied by 2) - 4,4
sum ...: ...
sum 12: 6,6 (just once)
... and count them all up, which gives 21 possible ways that give sum »greater than 6« (same way count sums of »at least 9«: 10 ways to do that), and thus getting \(\displaystyle P=\frac{\frac{10}{36}}{\frac{21}{36}}=\frac{10}{21}\). Is there any way to formalize this process (I assume that counting isn't math rigor ) :?:

d: same counting ... HELP

 

[soroban]

Re: Throwing two dice

Hello, courteous!

With "a pair of dice" problems, we usually list all the outcomes.

. . \(\displaystyle \begin{array}{cccccc}1,1&1,2&1,3&1,4&1,5&1,6\\2,1&2,2&2,3&2,4&2,5&2,6\\ 3,1&3,2&3,3& 3,4&3,5&3,6 \\ 4,1&4,2&4,3&4,4&4,5&4,6 \\ 5,1&5,2&5,3&5,4&5,5&5,6 \\ 6,1&6,2&6,3&6,4&6,5&6,6 \end{array}\)

We simultaneously throw two dice. What is the probability that:

a) At least one throw is divisible by 3 [SOLVED]

b) The sum of two throws is divisible by 3, given that the sum is divisible by 2

There are 18 outcomes which are divisible by 2.
Among them, 6 of them are divisible by 3.

\(\displaystyle \text{Therefore: }\(\text{div by 3 }|\text{ div by 2}) \;=\;\frac{6}{18} \;=\;\frac{1}{3}\)

c) The sum is at least 9, given that the sum is greater than 6

\(\displaystyle \text{Sum}> 6\!\;\;\begin{array}{c}1,6 \\ 2,5\quad2,6 \\ 3,4\quad3,5\quad\boxed{3,6} \\ 4,3\quad4,4\quad\boxed{4,5}\quad\boxed{4,6} \\ 5,2\quad5,3\quad\boxed{5,4}\quad\boxed{5,5}\quad\boxed{5,6} \\ 6,1\quad6,2\quad\boxed{6,3}\quad\boxed{6,4}\quad\boxed{6,5}\quad\boxed{6,6} \end{array}\;\;\boxed{\text{Sum}\geq 9}\)






\(\displaystyle \text{Therefore: }\(\text{at least 9 }|\text{ greater than 6}) \;=\;\frac{10}{21}\)

d) The difference between throws on each dice is 2, given that the sum is greater than 8.

\(\displaystyle \text{Sum}> 8:\;\;\begin{array}{c}3,6\\4,5\quad\boxed{4,6} \\ 5,4\quad5,5\quad5,6 \\ 6,3\quad\boxed{6,4}\quad6,5\quad6,6 \end{array}\;\;\boxed{\text{Diff of 2}}\)

\(\displaystyle \text{Therefore: }\(\text{diff of 2 }|\text{ sum}> 8) \:=\:\frac{2}{10} \:=\:\frac{1}{5}\)