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a) 2C1/10C4

b) 2C0/10C4

c) (2C1 + 2C2)/10C4

i am very confused...and i know the answers to these problems...i just dont know how to do them?

get it or nt?

no not really........the answer to the first is .8 so that is why i am confused

cant be

how?

10!
-------- sounds plausible?
2!(10-2)!

check my solution...shes sayin its 0.8 the answer

then it aint 45 lol

2 2 2 2
------- perchance?
10 9 8 7

combo and perms are still new to me :)

ha same

id have to draw up a tree or something and count them out lol

yeah...i am not sure if it is a combo though?
def probablility

its using combos

m 95% sure my methods right nelly

there are 45 ways to select 4 cans right? or did i mess that up too

10C4

10C4....what is that in longhand tho?

10!
--- ??
4!6!

10!/(6!4!)

210?

i got 120960?

yeah

so we have 210 ways to do this lol

great

how many ways can we have 2 tcans?

672

or is it easier to say how many ways to pick no tcans?
8 can ways then...

no no

pick one out of 2 cans is 2C1

but that doesnt gives us a total ways for tcans does it?

336

if there are 8 ways to choose 4 cans
8!
--- ??
4!4!

70?

8765432
--------
432.432
8765
-----
432
725 = 14*5 = 70

70/210 = 7/21 = 1/3; 33.33% ??

so the chance that 1 is at least a tcan could be 66' 2/3% does that seem reasonable?

33' 1/3% chance we got no tcans
66' 2/3% chance we got at least 1 tcan seems good to me

nope the anwser is .8 blah

.8 is 80%
which answer is .8? a,b or c?

whats the answers to b and c :)

b .333333
c. .66666

yay!! i was right :)

9!
---- = 3.3.2.7 = 9(14) = 126 ways to possible ways to
4!5! to choose at least 1 tcan

i am still very confused how to do the first as i get 122/210 = .53333

126 ways for 1 possible tcan
and we still have 70 ways for no tcans..; i wonder

56/210
.
---------
105/28.0000 ......not gonna be .8 lol