CPhill

Check me out here, Melody.....

We can choose any two of the 18 rolls to have the 3 occur.......and the probability that it occurs on any one roll is (1/6)......and the probaility that it doesn't occur is just (5/6)

So the probaility is

C(18, 2)* (1/6)^2 * (5/6)^16  = about 23%

I've been "rejuvenated" too, zegroes...!!!!

I forgot....Nauseated can be PTB.....

GL can be FTB   (Full-Time Bassist......)

GL, maybe we confer an auxiliary title on Nauseated.......PTB  .....Part-Time B*****d !!!

PTB-SOB......Mmmmm sort of has a ring to it, doesn't it ???

No biggie...I knew what you meant....!!!

Another good one GL......(we're really expanding here, rather than factoring).....(if I keep eating, I may expand forever !!! )

Good job, GL  !!!

LOL!!!

V_sphere  = (4/3)*pi* (r)^3  =  (4/3)(3.14)(8)^3 = about 2143.57 cubic inches

The total number of possible sets is given by C(10, 4)  = 210

And the ones we are interested in contain 2 white ones chosen from any 5 white ones and 2 black ones drawn from any 5 black ones

So we have

[ C(5, 2) * C(5, 2) ] / C(10, 4) =  [ 10 * 10 ] / 210  = 100 / 210  = about 47.6%