# Finding pmf for 4 variations of an urn problem.

I would like to test my understanding of distributions by verifying my answers to 4 deliberately related questions.

• Urn R has 5 R marbles.
• Urn G has 11 G marbles.
• Urn B has 4 B marbles.
• Random vector $$= F = \left[F_r\;F_g\;F_b\right]^T$$.
• # favorable outcomes vector $$= f = \left[f_1\;f_2\;f_3\right]^T = \left[f_r\;f_g\;f_b\right]^T$$.
• Probability of 1 favorable outcome vector $$= p = \left[p_1\;p_2\;p_3\right]^T = \left[p_r\;p_g\;p_b\right]^T$$.
• $$P\left[F = f\right] = P\left[F_r = 1, F_g = 2, F_b = 3\right]$$.

q1: A person successively draws 6 marbles, 1 @ a time; what is $$P\left[F = f\right]$$...

q1a: S.t. marbles are returned to the urn from which they were drawn, after every draw?

q1b: S.t. marbles are returned to a random urn, after every draw?

q2a: q1a if marbles are drawn 2 @ a time (w/o being drawn/returned from/to separate urns).

q2b: q1b if marbles are drawn 2 @ a time (w/o being drawn/returned from/to separate urns).

q3: A monster w/ 6 hands simultaneously draws 6 marbles; what is $$P\left[F = f\right]$$?

q4: Urns R & G are removed from game after being emptied into urn B...

q4a: q1a.

q4b: q2a.

q4c: q3.

a1a: $$P\left[F = f\right] = t!\prod_{k = 1}^c \frac{p_k^{f_k}}{f_k!} = \frac{6!(1/3)^{(1)}(1/3)^{(2)}(1/3)^{(3)}}{(1)!(2)!(3)!} = \frac{20}{243}$$.

a1b: Not a1a bc the draws are no longer independent (markov chain?).

a2a: $$P\left[F = f\right] = \begin{cases} t!\prod_{k = 1}^c \frac{p_k^{f_k}}{f_k!}, & \style{font-family:inherit}{\text{even f}} \\ 0, & \style{font-family:inherit}{\text{otherwise}} \end{cases} = \begin{cases} \frac{3!(1/3)^{f_1 + f_2 + f_3}}{f_1!f_2!f_3!}, & \style{font-family:inherit}{\text{even f}} \\ 0, & \style{font-family:inherit}{\text{otherwise}} \end{cases} = 0.$$

a2b: Not a2a bc the draws are no longer independent (markov chain?).

a3: $$P\left[F = f\right] = \frac{\left(\begin{smallmatrix} 5 \\ f_1 \end{smallmatrix}\right)\left(\begin{smallmatrix} 11 \\ f_2 \end{smallmatrix}\right)\left(\begin{smallmatrix} 4 \\ f_3 \end{smallmatrix}\right)}{\left(\begin{smallmatrix} 20 \\ t \end{smallmatrix}\right)} = \frac{\left(\begin{smallmatrix} 5 \\ 1 \end{smallmatrix}\right)\left(\begin{smallmatrix} 11 \\ 2 \end{smallmatrix}\right)\left(\begin{smallmatrix} 4 \\ 3 \end{smallmatrix}\right)}{\left(\begin{smallmatrix} 20 \\ 6 \end{smallmatrix}\right)} = \frac{55}{1938}$$.

a4a: $$P\left[F = f\right] = t!\prod_{k = 1}^c \frac{p_k^{f_k}}{f_k!} = \frac{6!(5/20)^{(1)}(11/20)^{(2)}(4/20)^{(3)}}{(1)!(2)!(3)!} = \frac{363}{10000}$$.

a4b: a4a.

a4c: a3.