# Probability for which die produced the 3?

Sorry for the title but I don't know how to put it in just a few words in a title

I have 3 dice: a d4, a d6 and a d8, where dk denotes a k-sided die.

I take a random die and I get a 3 ! What's the probability that I chose the d8 ?

Should I apply conditional probability ? Or it's just a simple 1/3 ? Or a simple 1/8 ?

So it might be needed a conditional probability but how?

P(A|B) = (P(A) + P(B) - P(A U B))/P(B)

That's how it might be: A -> a "3" from any of the dice - P(A) = 1/4+1/6+1/8 = (6+4+3)/24 = 13/24 B -> the d8 has been chosen - P(B) = 1/3 A U B -> a "3" from the d8 has appeared - P(A U B) = 1/8

So is it...

(13/34 + 1/3 - 1/8)/ 1/3 ?

• Is this homework? Please add the self-study tag and read its Wiki. What thoughts do you have on how to solve this problem? – Dave Feb 21 at 21:38
• Actually it's curiosity with friends that it translated in a huge debate, and now I want to know. :) I tried to give an answer. It might be a conditional probability and might be A : get a 3 from all the dice (1/4+1/6+1/8) B: choose one of the 3 dice (1/3) and A U B : get a "3" from a d8 (1/3) . But.. I don't know.. it doesn't add up? – salvob Feb 21 at 21:54
• I’d use Bayes: $P(d8 \vert 3) = \dfrac{P(3\vert d8) P(d8)}{P(3)}$. – Dave Feb 21 at 21:55
• P(d8) is the probability to get a d8 ? So 1/3 and p(3|d8)? – salvob Feb 21 at 22:07
• That’s the probability of rolling a 3 given that you select the d8. – Dave Feb 21 at 22:41

## 1 Answer

Firstly, your probability for A should be 13/72 because - hopefully it seems intuitively wrong to get more than half of the results be three, as implied by 13/24...

Secondly, the probability for the union means either occur, not both.

Thirdly, you should be calculating P(B|A). You condition on the three because this is what you observe. The probability for both is p(AB) = 1/3 × 1/8 = 1/24. This means the answer is P(B|A)=(1/24)/(13/72)=3/13