Question

A jar contains 8 green, 5 red, 4 blue, and 3 black marbles.once a marble is selected, it is not replaced. find each probability.
1) p(two green marbles)
2) p(two black marbles)
3) p(a red marble and then a blue marble)
4) p(a black marble then a green marble)
5) p(two marbles that are not red)
6) p(two marbles that are neither blue nor green)

7) a family has two children, both of whom are boys. what is the probability that the next child will be born a boy?

Answer 1

ready?

Answer 2

im new hear im pretty much lost

Answer 3

pick one
want to do the first one?

Answer 4

1)

Answer 5

pick one marble
how many are green?

Answer 6

i mean in total how many green ones are there?

Answer 7

8 green marbles

Answer 8

out of how many total?

Answer 9

12

Answer 10

actually the total is 20, the total includes the green ones

Answer 11

ok

Answer 12

therefore the probability you pick a green one on the fist try is \(\frac{8}{20}\)
is that much okay?

Answer 13

but it says they picked out 2 green marbles

Answer 14

yes i know, but we have to do it one step at a time

Answer 15

ok

Answer 16

first we pick a marble and the probability it is green is \(\frac{8}{20}\)
then we pick another marble. assuming that the first one we picked was green, how many green marbles are left in the bag?

Answer 17

7 out of 19

Answer 18

yes, there are 7 green ones left, and 19 total in the bag, so the probability that the second one is green given that the first one was also green is \(\frac{7}{19}\) what you said

Answer 19

now to compute the probability that BOTH are green, take those two numbers and multiply them

Answer 20

i.e. it is \(\frac{8}{20}\times \frac{7}{19}\)

Answer 21

thanks for your help but i really need to move on. i have alot of homework to do. thank you for your help. bye

Answer 22

yw
all the others are exactly the same idea

Answer 23

i still dont get it but anyways thanks for trying