There are 3 yellow, 5 red, 4 blue, and 8 green candies in a bag. once a candy is selected, it is not replaced. find the probability of:
p(two red candies)
p(two blue candies)
p(a yellow candy and then a blue candy)
p(a green candy and then a red candy)
p(two candies that are not green)
p(two candies that are neither blue nor green)
show calculations

probability=(desired outcomes)/(total possible outcomes)

so lets say we want to get 2 red candies
we can get out of 5 candies so
so desired outcomes=5 for first candy
find total
3+5+4+8=20
so first red candy=5/20=1/4
then second candy
it is 4/19 since 1 has been seletced already

so we have 1/4 and 1/19 so we multiply
1/4 times 1/19=1/76

2red=1/176

4 blue
get 2 blue
same thing
4/20 times 3/19=1/5 times 3/19=3/95
2blue=3/95

yellow and blue
yellos=3
blue=4
3/20 times 4/19=9/95
yellow then blue=9/95 

green and red
5 red8 green5/20 times 8/19=2/19
ggreen then red=2/19

not green
20-8=12
12=not green
12/20 times 11/19=33/95

not blue or green
blue+green=4+8=12
20-12=8
8 not green or blue
8/20 tiems 7/19=14/95

 
P(Two red candies)=1/76
P(Two blue candies)=3/95
P(A yellow candy and then a blue candy)=9/95
P(A green candy and then a red candy)=2/19
P(Two candies that are not green)=33/95
P(Two candies that are neither blue nor green)=14/95

The answer is the numbers between =5 and 5