In a bag, there are: 6 red marbles, 3 blue marbles, 4 yellow marbles, and 1 green marble. Without putting marbles back in the bag: a) What’s the probability of picking 3 red marbles?
b) What’s the probability of picking 2 green marbles?
c) What’s the probability of getting a blue marble first, and then a yellow marble?
d) What’s the probability of picking 1 red marble and 1 blue marble in any order?

Step-by-step explanation:

a) so initially there will be a total of 14 marbles in the bag since none have been removed yet. This means the probability of picking one red marble will be 6/14 or simplified, 3/7. However, the next time you pick a marble there will only be 13 marbles in the bag and only 5 red marbles, giving you a probability of 5/13. To get the next red marble, there will only be 12 marbles left in the bag with only 4 red marbles remaining. The probability of this is 4/12 or simplified, 1/3. In order to find the probability of picking 3 total red marbles, you need to multiply those probabilities together, so: (3/7)(5/13)(1/3) for an answer of 5/91 or a 0.0549% chance

b) I am not sure I fully understand how this question is possible if there is only one green marble and the marbles are not being replaced. My best guess would be a 0% chance since you cannot get 2 green marbles without replacement.

c) Again there are 14 marbles in the bag, that means the probability of picking a blue marble is 3/14, then, to find the probability of getting a yellow marble, you get 4/13, remember there is one less marble in the bag because you already took out a blue marble. Then simply multiply those probabilities together: (3/14)(4/13) for 6/91 probability or 0.0659% chance.

d) I am not entirely sure. If you want the probability of the red first then blue it is 9/91, and if it is blue then red, it is also 9/91.

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4