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?
3 Answers
$\begingroup$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:$$P(C)=\frac49\cdot \frac49=\frac{16}{81}$$
$P(A)=(\frac29)\cdot(\frac29)$
$P(B)=(\frac39)\cdot(\frac39)$
$P(C)=(\frac49)\cdot(\frac49)$
There is a very simple equation that everyone seems to forget to mention: $(x/t)^n$ where $x$ is the number of desired objects, $t$ is the total amount of objects, and $n$ is the number of times that you are drawing the object. For $P(A)$ it would be as follows: $(2/9)^2$; $P(B)$ would be: $(3/9)^2$ or $(1/3)^2$; etc.
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