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