Archman said:
12/51*11/50*10/49*9/48
and
((5*4*4)/(9*8*7)) * 3 + (5*4*3)/(9*8*7)
I wonder if that approach comes from non-HSC experience
![Wink ;) ;)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
hmm.. but it can be from 2U and/or ext2...
anyway the more ext1-ish way of doing these 2 questions would be:
"five cards are drawn at random from a pack of 52 playing cards. what is the probability that they are all from the same suit?"
Total possible outcomes: 52C5
Favourable outcomes: 13C5+13C5+13C5+13C5 = 4*13C5
"a bag contains 5 red balls and 4 white balls. 3 balls are drawn without replacement. find the prob. of drawing at least 2 red balls.."
Total possible outcomes: 9C3
Favourable outcomes: 5C2*4C1 (for when two are red) + 5C3 (when three are red)
you can also consider them as ordered selection, doesn't matter since it's probability.