# Probability with replacement marbles

Two marbles are drawn at random and with replacement from a box containing $$2$$ red, $$3$$ green, and $$4$$ blue marbles.
Let's define the following events:
A={two red marbles are drawn}
B={ two green marbles are drawn}
C={two blue marbles are drawn}.

Let's say i want to find the probability of A.
Since it's with replacement the first time i'm drawing, the probability would be $$\frac29$$ and the second time would also be $$\frac29$$ which would be $$\frac4{81}$$.
Is this the correct way of thinking this?

## 2 Answers

Yes, you are on a right track:

Total number of balls always remains $$9$$.

For event $$A$$:
There are $$2$$ Red balls, for both draws:
$$P(A)=\frac29\cdot \frac29=\frac4{81}$$ For event $$B$$:
There are $$3$$ Green Balls, for both draws: $$P(B)=\frac39\cdot\frac39=\frac{9}{81}$$ For event $$C$$:
There are 4 Blue Balls, for both draws: $$P(C)=\frac49\cdot \frac49=\frac{16}{81}$$

$$P(A)=(\frac29)\cdot(\frac29)$$
$$P(B)=(\frac39)\cdot(\frac39)$$
$$P(C)=(\frac49)\cdot(\frac49)$$