# Two fair dice (one red and one green) are rolled. What is the probability that the sum is 5, given that the green one is either 4 or 3?

Apr 1, 2017

Probability $= \frac{1}{6}$

#### Explanation:

Let $A$ be the event that the sum of the two dice is $5$
Let $B$ be the event that the green die is either a $3$ or $4$

Then we want $P \left(A | B\right)$ which we calculate using the conditional probability formula:

$P \left(A | B\right) = \frac{P \left(A \cap B\right)}{P \left(B\right)}$

Consider first $P \left(A \cap B\right)$ which we can calculate using

$P \left(A \cap B\right) = \frac{n \left(A \cap B\right)}{n \left(T\right)}$

Where, $n \left(T\right)$ is the total number of possible outcomes. As there are two dices then each has $6$ possible outcomes, making $n \left(T\right) = 36$.

And, $n \left(A \cap B\right)$ is the number of outcomes where the total is $5$ and the green die is a $3$ or a $4$.

If $G = 3 \implies R = 2$
If $G = 4 \implies R = 1$

And so $n \left(A \cap B\right) = 2$, and therefore we have:

$P \left(A \cap B\right) = \frac{2}{36} = \frac{1}{18}$

Now, let use calculate $P \left(B\right)$ in a similar fashion:

$P \left(B\right) = \frac{n \left(B\right)}{n \left(T\right)}$

Where $n \left(B\right)$ is the number of outcomes for which the green die is a $3$ or a $4$. For this outcome we have:

If $G = 3 \implies R = 1 , 2 , 3 , 4 , 5 , 6$
If $G = 4 \implies R = 1 , 2 , 3 , 4 , 5 , 6$

And so $n \left(B\right) = 12$, and s before $n \left(T\right) = 36$, and so:

$P \left(B\right) = \frac{12}{36} = \frac{1}{3}$

And so we can now calculate:

$P \left(A | B\right) = \frac{\frac{1}{18}}{\frac{1}{3}} = \frac{3}{18} = \frac{1}{6}$

Apr 1, 2017

See other answer for a "proper" discussion of how to evaluate conditional probabilities.
The explanation (below) is simply offered as an alternate, quick-and-dirty way of seeing this result.

#### Explanation:

If green is $\textcolor{g r e e n}{3}$ or $\textcolor{g r e e n}{4}$
and red is $\textcolor{red}{1}$, $\textcolor{red}{2}$, $\textcolor{red}{3}$, $\textcolor{red}{4}$, $\textcolor{red}{5}$, or $\textcolor{red}{6}$

{: (,,,color(green)("green"),), (,ul("Sum of"),ul(" | "),ul(color(green)3),ul(color(green)4)), (color(red)("red"),color(red)1," | ",4,color(magenta)5), (,color(red)2," | ", color(magenta)5,6), (,color(red)3," | ", 6,7), (,color(red)4," | ", 7,8), (,color(red)5," | ", 8,9), (,color(red)6," | ", 9,10) :}

As can be seen from the table:
$\textcolor{w h i t e}{\text{XXX}}$there are a total of $12$ possible outcomes
and
$\textcolor{w h i t e}{\text{XXX}}$ only $\textcolor{m a \ge n t a}{2}$ of those outcomes meet the requirement that the total be $5$

So the probability is $\frac{2}{12} = \frac{1}{6}$