Probability of R1 & R2: Solve w/ Tree Diagram

[crays]

Hi, i have this question
A bag contains 5 red balls, 10 blue balls and 15 green balls. Three balls are drawn from the bag one after another without replacement. The event R1 , R2 , B2 and G3 are defined as follows.

R1 - represents the event the first ball drawn is red.
R2 - represents the event the second ball drawn is red.
B2 - represents the event the second ball drawn is blue.
G3 - represents the event the third ball drawn is green.

Find
a) i) P(R1 $$\cap$$ R2)

for this , i drew a tree diagram, for it to be red for the first ball, it has to be 5/30 and for the second to be red it has to be 4/29 thus multiplying them both would give me the answer which is 2/87.

ii) P(R1 $$\cup$$ R2)

for this i thought of P(R1) + P (R2) - P(R1 $$\cap$$ R2) would solve the problem but the answer is incorrect and later i see, i don't really know what's the probability of R2 because it could be 5/29 or 4/29.

Please help.

 

[Dick]

You see the problem correctly. You could add up the separate cases i) first red, second not red, ii) first not red, second red and iii) both red. But it might be easier to notice that is equal to 1-(first not red)*(second not red).

 

[statdad]

You have this on another part of this site - look over there.