GingerAle

You didn’t set up any equations, so you don’t know what the hell is going on. You plopped two (2) combinations for multiplication with a comment about dividing.

In your plop, you removed the three (3) top students for selection (3C2), but then included one of the top students in the combination for non-top students. 

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To solve this:

Note that the “At Least” condition creates a mathematical “OR,” requiring the creation of two (2) sets: The first set has two of the top students; the second has all three top students. These sets are added giving the total number of sets with two (2) OR three (3) top students

$\dbinom{3}{2}*\dbinom{25}{2} = 900 \hspace {1em} | \hspace {2em} \text {Number of sets having two of the three top students.}\\ \dbinom{3}{3}*\dbinom{25}{1} = 25 \hspace {1em} | \hspace {2em} \text {Number of sets having all three of the top students.}\\ \text { }\\ 900 + 25 =925 \\ \text {Divide this by the total number of sets of four (4) from (28): } \binom{28}{4} = 20475 \\ \LARGE \rho \normalsize (r\geq 2)  = \dfrac{925}{20475} \rightarrow \dfrac{37}{819}$

GA

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It was Jack’s boo-boo. ...Jill kissed it, and made it all better. 

Solution:

Jack’s fist-half percentage is 75% with 15 successes,

Jack’s second-half percentage is 100% with 10 successes

Jack’s total percentage is 83.333% with 25 successes in 30 shots.

Jill’s total percentage is also 83.333% with 25 successes in 30 shots.

Jill’s fist-half percentage has to be less than 75% for 12 shots.

So, 0.75*12 = 9. Jill had 8 successes in 12 attempts (66.666%).

Then 25 - 8 = 17, so Jill had 17 successes in 18 attempts (94.44%).

Each of these percentages is less than Jack’s percentages for the first and second half.

Jill had (8) successes in the first half and (17) successes in the second half.

GA

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You may be decent at math, but logic ...

I'm not saying he wouldn't lie to me, but I don't think he would be untruthful.

Honestly, I lie all the time, but I’ll truthfully say, this is a contradiction!

GA

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Make sure there is a space after any and all < symbols ... else the forum’s server computer may eat it. 

...Non numeric values are delicious.  

GA 

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Here is a much easier way to solve this. The method does not use Hypergeometric selection

There are $(3^3) = 27$ arrangements of success. Divide the number of successes by the total number of sets giving $(3^3) / (nCr(9,3) =  \dfrac {9}{28} \approx 32.14\%$

This matches your calculation, Catmg. 

I’m glad you are paying attention to my presentations and train wrecks! 

If my train was loaded with toxic chemicals, I could wipe out a city. Or, in this case, at least send a student down the wrong track.   

The calculation is exactly twice the value I presented for the Hypergeometric solution.

The formula is correct, but I must have made a mistake in the calculator input. I always check complex equations three times, but still, it escapes me...

 I saved the input used for the calculation: (3!*nCr(6,2)*nCr(3,2))/((nCr(9,3)*nCr(6,3))

...I used a three (3) instead of a four (4) in the second binomial.

[My great uncle Cosmo was a locomotive engineer (and an electrical engineer), he would have flunked me for me for my train driving skills.]   

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Here’s a graphic, demonstrating the arrangements for the above solution.

$\hspace {.3em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.3em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ X & X & X \\ X & X & X \\ \end{array} } \right] \text {First arrangement of “R} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & R \\ X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & X \\ X & X & R \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & X & R \\ X & R & X \\ X & X & X \\ \end{array} } \right] \text {Fourth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .} \\ \, \\ \left[ {\begin{array}{ccc} X & X & X \\ X & X & X \\ R & R & R \\ \end{array} } \right] \text {27^{th} arrangement} \\ $

From this graphic, it’s easy to see the (3^3) = 27 arrangements of success.

GA

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WTFF!! Your solution may be free, but it’s also absurd and moronic. (Just because It’s Free doesn’t mean you should be smoking it!)  

The only white balls I see at the moment are in a specimen jar of formalin solution on a shelf in a cabinet. They’ve been in the specimen jar for over 21 years, and they’ve not changed much over that time.  The balls are kitty balls and were originally attached to my cat, known on the forum by his stage name, D.C. Thomas Copper.  I sometimes called him the balless wonder.  My cat was not amused by this, but my dog, Mr. Peabody, thought it quite funny. They were always busting each other’s balls. Mr. Peabody had the advantage though, because he knew where D.C. Thomas’ balls are.   

D.C. Copper is now 22-years-old, still alive and somewhat active; although he mostly cat-naps. Occasionally, after his dinner, he reviews the forum and offers suggestions for potential ball-busting troll posts.   ...Like this one 

GA

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