**Nota Bene:** This little bit of math is the keystone in my attempt here (still in draft status) to provide a sharp, clear articulation of the concept of relevance as that concept pertains to Relevant Logic. Here I invited members of the online Physics Forum to point out any mistakes in the math should I have made any. Since no one there pointed out any such mistakes, I will assume that the math is correct. Naturally, should it turn out that I did make mistakes in the math, I will be royally pissed. 🙂

This post belongs to the ‘I invite anyone and everyone to tear this to pieces, should they uncover any missteps’ category.

The subject here isn’t roses (this is an obscure allusion to a movie I saw in my childhood), but wormy and non-wormy red and yellow apples.

Wormy Red Apple Image courtesy of foodclipart.com

**First Situation: All Of The Red Apples Are Wormy; Only Some Of The Yellow Apples Are: ** Let’s start with the following situation: the pile of apples in the orchard comprises 16 apples. Eight of the apples are red. All of the red apples are wormy. Eight of the apples are yellow. Of these yellow apples, four are wormy. Let’s suppose that the DBA in the sky has assigned an identifying number (doubtlessly using the Apple Sequence Database Object in the sky) to each apple.

The Sample Space Ω =

Ω = { a1

^{rw}, a2^{rw}, a3^{rw}, a4^{rw}, a5^{rw}, a6^{rw}, a7^{rw}, a8^{rw}, a9^{yw}, a10^{yw}, a11^{yw}, a12^{yw}, a13^{yw}, a14^{yw}, a15^{yw}, a16^{yw}}

where a1…a*n* indicate the numbered apples, and the superscripts r, y, w, and ~~w~~ indicate a red apple, a yellow apple, a wormy apple, and a non-wormy apple respectively.

*E* is the event ‘a red apple gets drawn from the pile’ =

E= { a1^{rw}, a2^{rw}, a3^{rw}, a4^{rw}, a5^{rw}, a6^{rw}, a7^{rw}, a8^{rw}}

*F* is the event ‘a wormy apple gets drawn from the pile’ =

F= { a1^{rw}, a2^{rw}, a3^{rw}, a4^{rw}, a5^{rw}, a6^{rw}, a7^{rw}, a8^{rw},a9^{yw}, a10^{yw}, a11^{yw}, a12^{yw}}

And of course the intersection of *E* and *F*, *E* ∩ *F*, the set of apples that are *both* red *and* wormy =

{ a1

^{rw}, a2^{rw}, a3^{rw}, a4^{rw}, a5^{rw}, a6^{rw}, a7^{rw}, a8^{rw}}

I will be assuming that each apple in Ω has an equal probability of being drawn.

The conditional probability that the apple drawn from the pile is wormy given that it is red is 1, as you can see from the following steps:

P( F

|E ) = P(E∩F)/P(E)P(

E∩F) = |E∩F|/|Ω|=8/16 = 1/2P(E) =

|E| / |Ω| =8/16 = 1/2

So:

P(

E∩F)/P(E) = 1/2/1/2 = 1

So:

P( F

|E ) = 1

The conditional probability that an apple drawn from this pile is wormy given that it is red is 1.

Now P(F) = 12/16 = 3/4. Since P(E) = 1/2, P(E) * P(F) = 1/2 * 3/4 = 3/8. So in this case P(*E* **∩** *F*) != P(E) * P(F), since 1/2 != 3/8. But two distinct events are independent of one another if and only if

P(*E* **∩** *F*) = P(E) * P(F)

So in this case E and F are not independent events. The probability that the apple is wormy given that it is red increases to 1 from the 3/4 probability given just the draw from the pile, before observing whether the apple drawn is red or yellow. (Conversely, the probability that the apple is red given that it is wormy increases to 2/3 from 1/2 given just the draw from the pile.)

Today’s homage to Plato’s SYMPOSIUM is this image of a young boxer appearing on the cover of a computer book.

I have to admit that this is the only computer book I have ever bought just for its cover.

How can anyone get anything done, much less study computer science and ninth-grade math, with beauty like this walking the earth?

Update 11/12/2018: Made one revision for the sake of clarity.

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