# Using an unfair die, where rolling even is three times higher than of rolling odd, what is the probability of rolling a 5 given we've rolled an odd number?

$\frac{1}{3}$

#### Explanation:

We're given an unfair die (I'm assuming it's a loaded 6-sided die) such that the probability of even is three times more than the probability of odd.

What is the probability of rolling a 5 given that we've rolled an odd number.

Conditional probability is found using:

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

which says "The probability of event B, given event A, equals the probability of A intersect B divided by the probability of event A".

Event B is rolling a 5 and event A is the rolling of an odd number.

What is $P \left(A\right)$? It's $\frac{3}{12} = \frac{1}{4}$ (the probability of rolling a $1 , 3 , 5$ is divided by all the results possible, which is $1 , 2 , 2 , 2 , 3 , 4 , 4 , 4 , 5 , 6 , 6 , 6$ - we can list the evens three times to account for the loaded aspect of the die).

What is $P \left(A \cap B\right)$? It's #1/12 (the probability of rolling a 5 out of the 12 possibilities).

And so we get:

$P \left(B | A\right) = \frac{\frac{1}{12}}{\frac{1}{4}} = \frac{1}{12} \times \frac{4}{1} = \frac{4}{12} = \frac{1}{3}$