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Suppose you have two urns. At the beginning of the experiment, Urn 1 contains three yellow balls, three red balls, and three green balls. Also, Urn 2 contains one yellow ball, two green balls and four purple balls. Consider a two-stage experiment in which we randomly draw three balls from Urn 1 and move them to Urn 2, and then we randomly draw one ball from the updated Urn 2.

a.Define two events as follows:A = { Two Yellow balls and one green ball are moved to urn 2}

and

B ={A green ball is drawn from urn 2}

Find the probabilities of these two events.

b.Are A and B independent?

c.Find the probability that at least two of the balls moved from Urn 1 to Urn 2 were yellow, given that the ball drawn from Urn 2 was yellow. There is no need to simplify fraction

Work: For P(A) I did (3C2*3C0*3C1)/9C3 Is this correct? FYI 3C2= 3 choose 2 I am a little lost for P(B)? For B I need A to solve for it and I am lost on C as well.

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What you did to find $P(A)$ is okay: $$P(A)=\frac{\binom32\binom30\binom31}{\binom93}$$

For a fixed green ball located at first hand in urn1 (there are $3$ such balls) the probability to be drawn from urn2 at the second experiment is $\frac39\frac1{10}$. The first factor is the probability that it will be placed in urn1 at the first experiment and the second is the probability that - if this indeed happens - it will be drawn at the second experiment.

For a fixed green ball located originally in urn2 (there are $2$ such balls) the probability to be drawn at the second experiment is $\frac1{10}$.

This concerns $5$ mutually exclusive events and leads to:$$P(B)=3\cdot\frac39\frac1{10}+2\cdot\frac1{10}=\frac3{10}$$

Further it is not difficult to find that also $P(B\mid A)=\frac3{10}$ so that actually $P(B\mid A)=P(B)$.

This allows us to conclude that $A$ and $B$ are independent.

Let $X$ denote the number of yellow balls drawn at first experiment and let $E$ denote the event that a yellow ball is drawn at second experiment.

Then to be found is $P(X\geq2\mid E)=P(X=2\mid E)+P(X=3\mid E)$.

Here $P(X=i\mid E)P(E)=P(X=i\wedge E)=P(E\mid X=i)P(X=i)$ for $i=2,3$.

So finding $P(E)$ and $P(E\mid X=i)$ and $P(X=i)$ for $i=2,3$ is enough for finding $P(X\geq2\mid E)$.

Give it a try.

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