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For example, say I have am holding an object; $O$ is the event that the object is an orange, and $R$ is the event that the object is round.

$P(O|R) = \frac{P(R|O)}{P(R)}$

This is obviously incorrect, but it reads to me like: "the probability that the object is an orange given that it is round is the probability of an object being round given that it is an orange, divided by the probability that the object is round in general". Which seems reasonable.

Why is the above interpretation wrong?

Each One Chew
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1 Answers1

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“Philosophically” it’s wrong because the right hand side is about roundness — the only place Orange appears is as a given condition — so not sure how you can get any info on Orange on LHS from only data about roundness on RHS.

Conceptually, it’s just wrong.

From the definition of conditional probability:

$$P(O|R)= \frac{P(O\cap R)}{P(R)}$$

However:

$$ \frac{P(R|O)}{P(R)}= \frac{\frac{P(O\cap R)}{P(O)}}{P(R)}=\frac{P(R\cap O)}{P(R)P(O)}\neq P(O|R)$$

Annika
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  • I understand that this may be vague, but is there an informal english way to explain why it's incorrect? – Each One Chew Sep 29 '22 at 01:38
  • @EachOneChew — see my first paragraph. It’s not about why it’s mathematically wrong. – Annika Sep 29 '22 at 01:39
  • Makes sense, thanks! – Each One Chew Sep 29 '22 at 01:40
  • @EachOneChew If you understand what conditional probability is supposed to represent and draw a Venn diagram of this you’ll see you are doing something that has no sensible interpretation as a conditional probability – Annika Sep 29 '22 at 01:43