Probability math

Question

a container contains 7 yellow and 3 green balls. You draw  three balls at random. On each draw, if the three balls is yellow you set it  aside and if the ball is green you put it back in the container. Find the  probability that the third draw is green.

Answer 1

there are 4 ways to get green on the 3rd draw. Sum the probabilities of each way

Pr(Y,Y,G) = 7/10 x 7/9 x 3/8

Pr(Y,G,G) = 7/10x 3/9 x 3/9

Pr(G,Y,G) = 3/10 x 7/10 x 3/9

Pr(G,G,G) = 3/10 x 3/10 x 3/10

Add those 4 probabilities = Pr(Green on the 3rd draw)

Answer 2

Hi Alvik!

For this problem, let's look at the different possibilities we can have with this situation. Let G be green and Y be yellow. The possible situations are as follows for the third ball to be green:

Y Y G

Y G G

G G G

G Y G

So now let's look at the possibility of each of these events happening (note that if a yellow ball is drawn, it is taken out so the total amount of balls go down and the amount of yellows go down by one, while if a green ball is drawn it is put back):

Y Y G --> 7/10 * 6/9 * 3/8 = 126/720

Y G G --> 7/10 * 3/9 * 3/9 = 63/810

G G G --> 3/10 * 3/10 * 3/10 = 27/1000

G Y G --> 3/10 * 7/10 * 3/9 = 63/900

We now add all of these probabilities together since they are independent, and we get our answer:

The probability of the third draw being a green is approximately 0.5928, or exactly 1067/1800