# Probability question

1. Feb 23, 2009

### 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.

2. Feb 23, 2009

### HallsofIvy

Yes, that is correct.

For R2, there are two possiblities to consider:
1) The first ball drawn is red. The probability of that happening is 5/30 and then the probability the second ball drawn is also red is 4/29. The probability the first ball drawn is red and the second ball drawn is red is (5/30)(4/29)= 2/87 as you calculated in the first problem.

2) The first ball drawn is not red. The probability of that happening is 25/30 and then the probability the second ball drawn is red is 5/29. The probability the first ball drawn is not red and the second ball drawn is red is (25/30)(5/29)= 25/174.

The probability that a red ball is drawn second is P(R1 and R2)+ P((not R1) and R2)= 2/87+ 25/174= 29/174.

3. Feb 23, 2009

### crays

Then i guess my book is wrong for the answer, it says 9/29

4. Feb 23, 2009

Your book is correct. You have

$$P(R_1) = 1/6, P(R_2) = 1/6$$

and you have

$$P(R_1 \cap R_2) = 2/{87}$$

Now use

$$P(R_1 \cup R_2) = P(R_1) + P(R_2) - P(R_1 \cap R_2)$$