Abag contains 4 green 4 red 2 blue crayons, what is the probability of random choosing 2 red crayons without replacement

10 crayons in all.
P (red) = 4/10 or 2/5
P (2nd red) = 3/9 =1/3  since there are only 9 crayons left from which to choose and only three of them are red (since we already pulled one).

2/5  x 1/3 = 2/15  = P (2 red) with no replacement.

If there are 10 crayons in a bag with 4 green, 4 red and 2 blue you have :

4 chances out of 10 to get a green crayon, 4 chances out of 10 to get a red crayon and 2 chances out of 10 to get a blue one.

On the first try, you have 4 chances out of 10 to get a red crayon.

If you take this crayon out of the bag, it means that there are 9 crayons remaining with just three red crayon.

Then, you have 3 out of 9 chances to get a red crayon on the second try.

In order to know the probability to get a red crayon each try, we have to multiply each probability :

4/10 * 3/9 = 2/90

12/90 = 2/15

So you have 2 chance out of 15 to get 2 red crayons in a row 

Iam guessing c. or d.