Question

Box A contains 3 red and 4 green marbles. Box B contains 5 red and 2 green marbles. One marble is randomly selected from box A and its color is noted. If it is red, 2 reds are added to box B. If it is green, 2 greens are added to box B.
A marble is then selected from box B. Find the probability that the marble selected from box B is green.

Answer 1

two cases, first case is that the first marble chosen is red. that has probability \(\frac{3}{7}\)
then the probability that the next one chosen is green is \(\frac{2}{9}\) since box B now contains 9 marbles, of which only two are green
probability of both things occurring is\[\frac{3}{7}\times \frac{2}{9}=\frac{2}{21}\]

now you have to repeat the process, but this time assume the first marble chosen is green

Answer 2

once you get the probability that the first one is green and the second one is green, add it to the probability that the first one is red and the second one is green, that i computed above.
that will be your final answer

Answer 3

if you like, i can check your answer once you get it

Answer 4

I would like that but right now Im just taking what you said in :P Im still sot of confused

Answer 5

Sort*

Answer 6

if you have any question about any step let me know
the basic idea is that you have to compute two probabilities,
1) second is green if first is red ( i computed that above)
2) second is green if first is green (that is for you)

since those are the only two possibilities, and since those sets are clearly disjoint (you cannot have the first being red AND green) you add up the two answers

Answer 7

Is it 22/63?

Answer 8

hey satellite73 is helly the girl you help before i had to change my username because openstudy didnt let me log in can you help me again with my problems please

Answer 9

that is what i get, yes
so i guess you got it!

Answer 10

\[\frac{3}{7}\times\frac{2}{9}+\frac{4}{7}\times \frac{4}{9}\]