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?

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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?

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Answer 1 · 2017-04-06T04:53:00

$\frac{1}{3}$ $1/3$

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:

$P (B ∣ A) = \frac{P (A \cap B)}{P (A)}$ $P(B|A)=(P (A nn B))/(P(A))$

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

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

And so we get:

$P (B ∣ A) = \frac{\frac{1}{12}}{\frac{1}{4}} = \frac{1}{12} \times \frac{4}{1} = \frac{4}{12} = \frac{1}{3}$ $P(B|A)=(1/12)/(1/4)=1/12xx4/1=4/12=1/3$

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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?

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