GingerAle

Actually, I didn’t get it wrong.  My comment has nothing to do with Geno’s avatar. It is simply a metaphor I use when something is about half completed.

“I’ve completed the hippo and I’m working on the potamus,” is my typical reply when a patron asks about the progress on a commissioned portrait. One time I responded this way to a patron who asked about my progress on his wife’s life-size portrait. He was rather offended and I had to explain I wasn’t referring to his wife as hippopotamus.

Because of my troll-chimp nature, I added that she is rather hippy and does have the largest potamus I’ve seen in a long time.  He was awkwardly quiet for a long time, but he didn’t cancel the commission until I said that I bought the economical 37-liter vats of oils.  It’s a pity that some people just do not have a sense of humor.

Geno’s avatar does look somewhat like a hippopotamus when displayed in thumbnail size.  I kinda, sorta figured a blarney bag would come along and point out the obvious error of my ways.  He did. And he says, “All rhinoceroses are endangered.” Why, I had no idea!!! That’s a good thing to know. I’ll cut back on my use of rhinoceros horn. I wouldn’t want to be a contributor to the extinction of rhinoceroses – just blarney bags.

Geno, you have the hippo but you are missing the potamus.

 (Using subscripted variables makes it easier to present and keep track of the relations.)

\(\small \text {Let A1 = {pass exam 1}, A2 = {pass exam 2}, A3 = {pass exam 3}, and A = {passes all three exams}. }\\\)

\(\text{Because }\ A^c_2\subset A^c, \; A^c_2\cap Ac=A^c_2 \\ \mathbb{P}(A^c_2) = \mathbb{P}(A^c_2|A^c_1)\; \cdot\; \mathbb{P}(A^c_1 +\mathbb{P}(A^c_2|A_1) \; \cdot\; \mathbb{P}(A_1) \leftarrow \small \text{Apply Total Probability Formula}\\ = (0 \cdot 0.1) + (0.2 \cdot 0.9) = 0.18 \\ \text {Then }\\ \mathbb{P}(A^c_2|A_c) = \dfrac {\mathbb{P}(A^c_2 \cap A^c)} {\mathbb{P}(A^c)}\\ = \dfrac {0.18}{0.496} = 0.363\)

GA

The solution should resolve to $39.84.

Melody and the obsequious Morgan Tud present two methods to solve this question. Melody’s method is straightforward and easier to understand. Tud derives a quadratic equation from the two relations. It’s a brilliant method, but not intuitive for most persons trying to solve these types of questions. 

I have a crystal ball –compliments of Morgan Tud. I’ve not yet figured out the mystery to make it work. However, I do see very pretty things in the orb –usually my garden flowers or my own face. Sometimes it’s both at the same time –it’s really beautiful then.           

This is (as JB calls it),”slop at the top!” Of course, Mr. BB mixed this slop with blarney; I would expect nothing less from the Blarney Master.   Mr. BB’s presentations are always s slop, but the contrast is quite sharp when you compare it to Heureka’s masterful presentation.

Let’s see:  you “don’t know what the formula is supposed to calculate!”  But you know it is about calculating the ‘escape velocity’” (it’s impressive that you know that).  “But you don’t need it here!”  That’s impressive too, because it’s a monumental metric for measuring Batshit Stupid!  

The asker is supposed to ignore the formula that gives the Schwarzschild radius, but know “…that if the Earth was squeezed into a black hole, the black hole would have a radius of 9 millimeters!”

How is he supposed to know that?  Oh, of course, ask your computer: it’s connected to the internet, and everything read on there is true; that’s a fact; you can read about on the internet!    

After he finds out (via magic incantation) that it is “9 millimeters!”  He’s supposed to know “The ‘photon sphere’ or the orbit of light would be: 1.5 x 9 =13.5 mm from the center of the black hole. Or, 13.5 - 9 =4.5mm from the ‘event horizon’.” He knows this the same way he knows the first part . . .  He asks his computer. No need for formulas or derivations. That’s just a waste of time!   

None of this matters because this is not the question.  The photon sphere is irrelevant here. Only the escape velocity is relevant. 

“That is the best I can do!.”

Oh surely not! Ask your computer this question:  What would the minimum orbital radius be if the partial were an electron?  

I can hardly wait!  I’ll be biding my time playing Monopoly with my dog and cat. It’s the cat’s turn to be the banker so it should be an interesting game!

What? I was on an hour ago.  Didn’t you see my brilliant post?

Logging on is a pain in my chimp rump --I mean my delicate heinie. The security certificates corrupt and I have to reinstall them. They are probably corrupted from all the allergy-causing blarney that passes through them.

JB, if you are going to imitate me in your troll posts at least target posts with wrong answers.  Mr BB gave the right answer. Least wise it exactly matches the answer by Presh Talwalkar here:

Tiggsy, your presentation is amazing. This clearly indicates all the mathematical relations, including the 10% “adjustment” for double counts.

Here’s a condensed presentation

\(\text{Prb = }\left(\dfrac{\binom{3}{2}\binom{17}{1} + \binom{3}{3}}{\binom{20}{3}} * \dfrac{9}{10}\right) + \dfrac{1}{10} \approx 0.1411 \text{ or } 14.11%\)

Melody’s Venn diagram also makes clear why the adjustment only affects the intersection of the two events and not the events separately.

At first blink, this question seems straightforward, and relatively simple to solve. However, it has subtleties that require the mathematician to have practiced skills in combinatorics and statistics to fish out the nuances that affect the final counts. Melody and Tiggsy obviously have these skills. 

Someone previously asked this question on here:

Hi Melody:

Actually the difference in our solutions is significant; it’s not from rounding errors.

The main reason is 

\(\dfrac{52}{1140}+\dfrac{1}{10}\underbrace {- \dfrac{52}{11400}}_{\text{This Term}}\\\)

^{(My computer is not rendering any Latex. I will correct this display, if needed)}

This subtracts 10% of the (2 of 3) ball wins.

I had initially thought of this but dismissed it because of the either/or nature of the win.

On reflection, I can see it does count the win twice when both the super ball and the white balls are a match. It should count one or the other but not both. The problem, now, is by subtracting 52/11400, it subtracts both wins.  So, what should the value be?

My quick answer is half that (26/11400).The reasoning is by subtracting half, then one or the other is subtracted but not both.

What thinkest thee, Dame Melody?

More comments:

I agree with the 52/1140, I neglected to include the 3 of 3 win in my equation, because I was patting my chimp head for figuring out Lancelot Link’s convoluted formula for counting a  Z subset of  matches of R in a  N choose R combination.  

Here are the adjusted numbers in the same format
6)  Add (1 / 1140) to (51 / 1140) giving (52 / 1140) for when all three numbers match.

Calculate number of times both the super ball and two of the three balls match
(52 / 1140)  * (1 / 10) = 52/11400      <---  ½ of this = (26 / 11400)

Subtract half of matches when both the super ball and two of the three balls match (This counts one or the other but not both.)

Calculate the harmonic average

H = 3 / ((52 / 1140) + (1 / 10) - (26 / 11400))  = 20.930232558

Equalized probability of each event (1 / 20.9023) * 3 (for the three events)
= (3 / 20.9023)
= (43 / 300)
= 0.143333…

Note: doing it this way is much easier (This is Melody’s method)

 ((52 / 1140) + (1 / 10) - (26 / 11400)) = (0.143333… )

Over all probability 14.33%

(Only the final calculation is rounded)

This question is interesting, but it seems very advanced.
Do you understand this math, Victoria?  

Comments and criticisms from mathematicians (and articulate trolls) are welcome.

GA

Hi Melody