Question

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?

Answer 1

`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(B|A)=(P (A nn B))/(P(A))`

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(A)`? It's `3/12=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(AnnB)`? It's #1/12 (the probability of rolling a 5 out of the 12 possibilities).

And so we get:

`P(B|A)=(1/12)/(1/4)=1/12xx4/1=4/12=1/3`