kay tears off labels of all 10 soup cans her mother knows there are 2 tomato and 8 vegetable she selects 4 at random. what is the probability a. exactly one of the 4 cans is tomato b. none of the 4 cans is tomato c. at least one of the 4 is tomato?

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kay tears off labels of all 10 soup cans her mother knows there are 2 tomato and 8 vegetable she selects 4 at random. what is the probability a. exactly one of the 4 cans is tomato b. none of the 4 cans is tomato c. at least one of the 4 is tomato?

Mathematics
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a) 2C1/10C4
b) 2C0/10C4
c) (2C1 + 2C2)/10C4

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Other answers:

i am very confused...and i know the answers to these problems...i just dont know how to do them?
on top of the fraction,the numerator is the no of ways o selecting 1 tomato can out of a possible 2 , and the denominator is the total no. of selections ie 10C4, selecting any 4 out of 10 cans
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?
10 9 8 7 6 5 4 3 2 ---------------- 4 3 2 6 5 4 3 2 10 9 8 7 -------- 4 3 2 5 3 2 7 = 15(14) = 210 right?
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?
10C4 is the total number of ways to choose 4 cans.... 8C4 is the total number of ways to choose no tcans.... casue we removed them from the equation.... % = part/total % = 8c4/10c4 for no tcans 100- no tcan% = at least 1 tcan.... is my reasoning
a
whats the answers to b and c :)
b .333333 c. .66666
yay!! i was right :)
exactly one of the 4 cans is tomato lets see if we take away one tcan; just for the sake of argument; we can do this then
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
126-70 would give us the number of ways for 1 tcan right? 126 70 ---- 56 ways that at least 1 tcan is in it... i hope lol
56/210 . --------- 105/28.0000 ......not gonna be .8 lol

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