Hi all. I'm struggling with a problem and I'm hoping there's a quick solution rather than having to add 80 equations together. It has been a few 15 or so years still I last studying probability and I've spent a couple of days trying to work out which combination equations will save me time. I'm looking more for the method than the answer and I've altered it so it could be translated to a poker or lottery question.
There are 52 balls. 18 of them marked a, 13 marked b, 4 c, 3 d,3 e,3 f, 2 g,1 h, 1 i, 1 j, 1 k, 1 l, 1 m.
(aaaaaaaaaaaaaaaaaabbbbbbbbbbbbbccccdddeeefffgghijklm)
What is the probability of having at least 5 different letters drawn if 7 balls are drawn?
so there are 52C7 or 133784560 different ways the 7 balls can be drawn
I was thinking the way to attack it would be
1-(Fail/(52C7)) (*100 to get a percentage)
where failing has 80 different equations ranging from 7a, 7b, 6a,...4a, 4b, 4c,..., (3a+2b), (3a+2c),...(3a+2g), (3b+2a),...,(2b+2d+2f),...(2e+2f+2g)
where the chance of getting 7a would be 18/52*17/51*16/50*15/49*14/48*13/47*12/46
and 2b+2d+2f would be 13/52*12/51*3/50*2/49*3/48*2/47*33/46
Any help getting this out of my head would be appreciated
There are 52 balls. 18 of them marked a, 13 marked b, 4 c, 3 d,3 e,3 f, 2 g,1 h, 1 i, 1 j, 1 k, 1 l, 1 m.
(aaaaaaaaaaaaaaaaaabbbbbbbbbbbbbccccdddeeefffgghijklm)
What is the probability of having at least 5 different letters drawn if 7 balls are drawn?
so there are 52C7 or 133784560 different ways the 7 balls can be drawn
I was thinking the way to attack it would be
1-(Fail/(52C7)) (*100 to get a percentage)
where failing has 80 different equations ranging from 7a, 7b, 6a,...4a, 4b, 4c,..., (3a+2b), (3a+2c),...(3a+2g), (3b+2a),...,(2b+2d+2f),...(2e+2f+2g)
where the chance of getting 7a would be 18/52*17/51*16/50*15/49*14/48*13/47*12/46
and 2b+2d+2f would be 13/52*12/51*3/50*2/49*3/48*2/47*33/46
Any help getting this out of my head would be appreciated
