There are 10 red and 20 yellow balls. drawing 3 balls without replacement, what is the chance the first is red and the next two are yellow?

10 + 20 = 30 total balls.

Picking red = 10/30 = 1/3

After picking one ball there are 29 balls left.

Picking a yellow next is 20/29

Then there are 28 balls left and 19 yellow left, so picking another yellow would be 19/28

The probability of picking red first then two yellow = 1/3 x 20/29 x 19/28 = 95/609

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3

idk

step-by-step explanation:

use that smart brain