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?

- anonymous

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- anonymous

a) 2C1/10C4

- anonymous

b) 2C0/10C4

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

get it or nt?

- anonymous

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

- anonymous

cant be

- anonymous

how?

- amistre64

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

- anonymous

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

- amistre64

then it aint 45 lol

- amistre64

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

- amistre64

combo and perms are still new to me :)

- anonymous

ha same

- amistre64

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

- anonymous

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

- anonymous

its using combos

- anonymous

m 95% sure my methods right nelly

- amistre64

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

- anonymous

10C4

- amistre64

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

- amistre64

10!
--- ??
4!6!

- anonymous

10!/(6!4!)

- amistre64

210?

- anonymous

i got 120960?

- amistre64

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?

- anonymous

yeah

- amistre64

so we have 210 ways to do this lol

- anonymous

great

- amistre64

how many ways can we have 2 tcans?

- anonymous

672

- amistre64

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

- anonymous

no no

- anonymous

pick one out of 2 cans is 2C1

- amistre64

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

- anonymous

336

- amistre64

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

- amistre64

70?

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

nope the anwser is .8 blah

- amistre64

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

- amistre64

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

- anonymous

a

- amistre64

whats the answers to b and c :)

- anonymous

b .333333
c. .66666

- amistre64

yay!! i was right :)

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

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