Please help me understand conditional probability in relation to dice If 2 fair dice are rolled, what is the probability that the sum is 6, given both dice show odd numbers?
When the sum is 6, there combinations are: (1,5),(2,4),(3,3),(4,2),(5,1).  Probability = 5/36
Probability of 2 odd numbers is 1/2.
What I'm not sure about is when we work out Pr(sum is 6|both numbers odd).  I initially thought that the top line was Pr(sum is 6), but then the result would be 5/2.  This can't be right.
So, does that mean that we only choose combinations which have odd pairs?  So, the answer would be 3/2?
Still makes no sense as the number is greater than 1???
Can somebody shed some light on this for me?
 A: What you're trying to compute is the probability that two dice sum to $6$, given that both dice have odd values. Note that there are three ways for pairs of dice with odd values to sum to six, namely $(1,5),(3,3),(5,1)$. There are three odd values each die can take, so there are nine ways for pairs of dice to both have odd values. Thus the probability you are looking for is $3/9=1/3$.
A: With conditional probabilities, the formal notation is often helpful, at least as an accounting device to keep things straight. In more complicated problems, it becomes more or less essential. So let's do things quite formally. 
Let $B$ be the event "both dice are odd," and let $S_6$ be the event "the sum is $6$."  We want
$$\Pr(S_6|B).$$
Now we use the formula
$$\Pr(S_6|B)\Pr(B)=\Pr(S_6\cap B),\tag{$1$}$$ 
or another version of the same thing.  The formula tells us that everything will be finished once we know $\Pr(B)$ and $\Pr(S_6\cap B)$. 
Calculating $\Pr(B)$ is easy: the answer is $\frac{9}{36}$.
To calculate $\Pr(S_6\cap B)$, note that the results on the two dice (coloured red and green say) can be represented as an ordered pair $(x,y)$ where $1\le x\le 6$, $1\le y\le 6$. 
These $36$ ordered pairs are all equally likely. How many have sum $6$ and are both odd? List them all: $(1,5)$, $(3,3)$, $(5,1)$. We conclude that $\Pr(S_6\cap B)=\frac{3}{36}$. 
Now use Equation $(1)$. The required conditional probability is $\dfrac{\frac{3}{36}}{\frac{9}{36}}$.
A: There are $36$ outcomes, of which, only the $9$ in red come from both dice showing odd numbers:
$$
\begin{array}{c|cccccc}
&1&2&3&4&5&6\\ \hline
1&\color{#C00000}{2}&3&\color{#C00000}{4}&5&\color{#C00000}{6}&7\\
2&3&4&5&6&7&8\\
3&\color{#C00000}{4}&5&\color{#C00000}{6}&7&\color{#C00000}{8}&9\\
4&5&6&7&8&9&10\\
5&\color{#C00000}{6}&7&\color{#C00000}{8}&9&\color{#C00000}{10}&11\\
6&7&8&9&10&11&12
\end{array}
$$
Each outcome is equally likely, so each of the red outcomes is also equally likely. Of the $9$ red outcomes, there are $3$ that total $6$. This means that, given that each die shows an odd number (the red outcomes), there is a $3/9=1/3$ probability that the sum is a $6$.
A: Soln : The  elements in sample space is $n(s) = 36$
Let  A = event  that  sum  is  $6 = \{ (15), (24), (33), (42), (51) \} $
$n(A) = 5$ so, $P(A) = 5/36$
and  B = event  that  both dice show odd numbers  is
           $= \{ (11), (13), (15), (31), (33), (35), (51). (53), (55) \} $         $n(B) = 11$ so, $P(B) = 9/36$
Now,  $A∩B = \{(15), (33), (51)\}$
$P(A \cap B) = 3/ 36 $
We  have  to  find  $P(A/B) = \{ P(A \cap B) \}/P(B) = (3/36)/(9/36) = 1/3$
