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}


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.